En mathématiques , les coefficients binomiaux de Gauss (également appelés coefficients de Gauss, polynômes de Gauss ou coefficients q- binomiaux ) sont des q-analogues des coefficients binomiaux . Le coefficient binomial de gauss, noté Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math> \binom nk_q}
</mo></mrow><mfrac linethickness="0"><mi>
(
n
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q
{\displaystyle {\binom {n}{k}}_{q}}
(
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q
{\displaystyle {\binom {n}{k}}_{q}}
(
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k
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q
{\displaystyle {\binom {n}{k}}_{q}}
(
n
k
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q
{\displaystyle {\binom {n}{k}}_{q}}
(
n
k
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q
{\displaystyle {\binom {n}{k}}_{q}}
(
n
k
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q
{\displaystyle {\binom {n}{k}}_{q}}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mo> <math>\begin{bmatrix}n\\ k\end{bmatrix}_q}
</mo><mtable rowspacing="4pt"><mtr><mtd><mi>
[
n
k
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q
{\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}
[
n
k
]
q
{\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}
[
n
k
]
q
{\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}
[
n
k
]
q
{\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}
[
n
k
]
q
{\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}
[
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k
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q
{\displaystyle {\begin{bmatrix}n\\k\end{bmatrix}}_{q}}
q avec des coefficients entiers qui lorsque la valeur de q est fixé à une puissance d'un nombre premier, compte le nombre de sous-espaces de dimension k dans un espace vectoriel de dimension n sur un corps fini à q éléments.
Les coefficients binomiaux de Gauss sont définis par
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = \begin{cases} \frac{(1-q^m)(1-q^{m-1})\cdots(1-q^{m-r+1})} {(1-q)(1-q^2)\cdots(1-q^r)} & r \le m \\ 0 & r>m \end{cases}}
</mo></mrow><mfrac linethickness="0"><mi>
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
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{\displaystyle {m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}}
où m et r sont des entiers naturels. Pour r = 0des produits sur l'ensemble vides . Bien que la formule dans le premier cas semble impliquer une fonction rationnelle , elle désigne en fait un polynôme, car la division est exacte dans Z [ q ] . Notez que la formule peut être appliquée pour r = m + 11 − q 0 = 0 dans le numérateur, conformément au second cas (pour r encore plus grand le facteur 0 reste présent dans le numérateur, mais ses autres facteurs impliqueraient des puissances négatives de q , donc l’énonciation explicite du second cas est préférable). Tous les facteurs du numérateur et du dénominateur sont divisibles par 1 − q , en tant que quotient un q-nombre définit comme suit :
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo stretchy="false"> <math>[k]_q=\sum_{0\leq i<k}q^i=1+q+q^2+\cdots+q^{k-1}= \begin{cases} \frac{1-q^k}{1-q} & \text{for} & q \neq 1 \\ k & \text{for} & q = 1 \end{cases},}
</mo><mi>
[
k
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,
{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
[
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
[
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
[
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
[
k
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=
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1
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1
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k
for
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
[
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=
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1
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{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}
en divisant ces facteurs, on obtient la formule équivalente
Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q=\frac{[m]_q[m-1]_q\cdots[m-r+1]_q}{[1]_q[2]_q\cdots[r]_q}\quad(r\leq m),}
</mo></mrow><mfrac linethickness="0"><mi>
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
(
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
(
m
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q
=
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}
ce qui met rend évident le fait que remplacer q par 1 dans Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr_q}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr.}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
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.
{\displaystyle {\tbinom {m}{r}}.}
(
m
r
)
.
{\displaystyle {\tbinom {m}{r}}.}
(
m
r
)
.
{\displaystyle {\tbinom {m}{r}}.}
(
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.
{\displaystyle {\tbinom {m}{r}}.}
(
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{\displaystyle {\tbinom {m}{r}}.}
{\displaystyle }
(
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.
{\displaystyle {\tbinom {m}{r}}.}
(
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)
.
{\displaystyle {\tbinom {m}{r}}.}
factorielle Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo stretchy="false"> <math>[n]_q!=[1]_q[2]_q\cdots[n]_q}
</mo><mi>
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{\displaystyle [n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}}
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{\displaystyle [n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}}
Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q=\frac{[m]_q!}{[r]_q!\,[m-r]_q!}\quad(r\leq m),}
</mo></mrow><mfrac linethickness="0"><mi>
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q
!
[
r
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q
!
[
m
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r
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q
!
(
r
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m
)
,
{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
(
m
r
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q
=
[
m
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q
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[
r
]
q
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[
m
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(
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m
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
(
m
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m
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[
r
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q
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m
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(
r
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
(
m
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q
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m
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
(
m
r
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[
m
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m
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
(
m
r
)
q
=
[
m
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[
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m
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
(
m
r
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q
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[
m
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[
r
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m
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r
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{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}
une forme compacte (souvent considérée comme une simple définition), qui cache cependant la présence de nombreux facteurs communs au numérateur et au dénominateur. Cette forme rend évidente la symétrie Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr_q=\tbinom m{m-r}_q}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
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r
)
q
=
(
m
m
−
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)
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{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
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r
)
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(
m
m
−
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)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
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)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
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)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
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q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
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)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
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)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
(
m
r
)
q
=
(
m
m
−
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}}
r ≤ m
Contrairement au coefficient binomial ordinaire, le coefficient binominal de gauss prend des valeurs finies lorsque Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math>m\rightarrow \infty}
</mi><mo stretchy="false">
m
→
∞
{\displaystyle m\rightarrow \infty }
m
→
∞
{\displaystyle m\rightarrow \infty }
m
→
∞
{\displaystyle m\rightarrow \infty }
m
→
∞
{\displaystyle m\rightarrow \infty }
q | < 1):
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{\infty \choose r}_q = \lim_{m\rightarrow \infty} {m \choose r}_q = \frac{1}{[r]_q!\,(1-q)^r}}
</mo></mrow><mfrac linethickness="0"><mi mathvariant="normal">
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
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→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
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(
m
r
)
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=
1
[
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]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
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)
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=
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m
→
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(
m
r
)
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=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
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(
m
r
)
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=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
(
∞
r
)
q
=
lim
m
→
∞
(
m
r
)
q
=
1
[
r
]
q
!
(
1
−
q
)
r
{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{0 \choose 0}_q = {1 \choose 0}_q = 1}
</mo></mrow><mfrac linethickness="0"><mn>
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
(
0
0
)
q
=
(
1
0
)
q
=
1
{\displaystyle {0 \choose 0}_{q}={1 \choose 0}_{q}=1}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{1 \choose 1}_q = \frac{1-q}{1-q}=1}
</mo></mrow><mfrac linethickness="0"><mn>
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
(
1
1
)
q
=
1
−
q
1
−
q
=
1
{\displaystyle {1 \choose 1}_{q}={\frac {1-q}{1-q}}=1}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{2 \choose 1}_q = \frac{1-q^2}{1-q}=1+q}
</mo></mrow><mfrac linethickness="0"><mn>
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
(
2
1
)
q
=
1
−
q
2
1
−
q
=
1
+
q
{\displaystyle {2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{3 \choose 1}_q = \frac{1-q^3}{1-q}=1+q+q^2}
</mo></mrow><mfrac linethickness="0"><mn>
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
(
3
1
)
q
=
1
−
q
3
1
−
q
=
1
+
q
+
q
2
{\displaystyle {3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}}
Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{3 \choose 2}_q = \frac{(1-q^3)(1-q^2)}{(1-q)(1-q^2)}=1+q+q^2}
</mo></mrow><mfrac linethickness="0"><mn>
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
q
2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
q
2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
q
2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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(
1
−
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2
)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
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(
1
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2
)
(
1
−
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)
(
1
−
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)
=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
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(
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2
)
(
1
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(
1
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2
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=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
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(
1
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(
1
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=
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
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(
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(
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{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
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=
(
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(
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{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
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=
(
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(
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{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
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(
1
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=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
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(
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=
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+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
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=
(
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(
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=
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+
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{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
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(
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=
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+
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{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
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=
(
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(
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
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=
(
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−
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3
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(
1
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)
(
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)
(
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−
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2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
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=
(
1
−
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(
1
−
q
2
)
(
1
−
q
)
(
1
−
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2
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=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
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=
(
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(
1
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2
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(
1
−
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(
1
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=
1
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+
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{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
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q
=
(
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(
1
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2
)
(
1
−
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)
(
1
−
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2
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=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
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−
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(
1
−
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2
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(
1
−
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)
(
1
−
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2
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=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
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(
1
−
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2
)
(
1
−
q
)
(
1
−
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=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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(
1
−
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2
)
(
1
−
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)
(
1
−
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2
)
=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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(
1
−
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2
)
(
1
−
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(
1
−
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=
1
+
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+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
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3
)
(
1
−
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2
)
(
1
−
q
)
(
1
−
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2
)
=
1
+
q
+
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2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
(
3
2
)
q
=
(
1
−
q
3
)
(
1
−
q
2
)
(
1
−
q
)
(
1
−
q
2
)
=
1
+
q
+
q
2
{\displaystyle {3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{4 \choose 2}_q = \frac{(1-q^4)(1-q^3)}{(1-q)(1-q^2)}=(1+q^2)(1+q+q^2)=1+q+2q^2+q^3+q^4}
</mo></mrow><mfrac linethickness="0"><mn>
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
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)
(
1
−
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2
)
=
(
1
+
q
2
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(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
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)
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1
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1
+
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+
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=
1
+
q
+
2
q
2
+
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3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
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q
=
(
1
−
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4
)
(
1
−
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(
1
−
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1
+
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=
1
+
q
+
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q
2
+
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3
+
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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q
=
(
1
−
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4
)
(
1
−
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(
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)
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1
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+
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+
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2
+
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
(
1
−
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(
1
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(
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1
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+
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2
+
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+
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
(
1
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(
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(
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1
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1
+
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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1
−
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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1
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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=
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
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q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
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2
)
(
1
+
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+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
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2
)
(
1
+
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+
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2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
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2
)
(
1
+
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+
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2
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=
1
+
q
+
2
q
2
+
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3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
q
)
(
1
−
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)
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(
1
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(
1
+
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=
1
+
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+
2
q
2
+
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+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
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)
(
1
−
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2
)
=
(
1
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2
)
(
1
+
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+
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)
=
1
+
q
+
2
q
2
+
q
3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
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2
)
(
1
+
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+
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)
=
1
+
q
+
2
q
2
+
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3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
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)
(
1
−
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2
)
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(
1
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(
1
+
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+
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=
1
+
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+
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q
2
+
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3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
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3
)
(
1
−
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)
(
1
−
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2
)
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(
1
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2
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(
1
+
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+
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=
1
+
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q
2
+
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3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
q
2
)
(
1
+
q
+
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2
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=
1
+
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+
2
q
2
+
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3
+
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4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
q
2
)
(
1
+
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+
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2
)
=
1
+
q
+
2
q
2
+
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3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
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2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
(
1
−
q
4
)
(
1
−
q
3
)
(
1
−
q
)
(
1
−
q
2
)
=
(
1
+
q
2
)
(
1
+
q
+
q
2
)
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}}
Description combinatoire
modifier
Au lieu de ces expressions algébriques, on peut également donner une définition combinatoire des coefficients binomiaux de gauss. Le coefficient binomial ordinaire Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
)
{\displaystyle {\tbinom {m}{r}}}
(
m
r
)
{\displaystyle {\tbinom {m}{r}}}
(
m
r
)
{\displaystyle {\tbinom {m}{r}}}
(
m
r
)
{\displaystyle {\tbinom {m}{r}}}
(
m
r
)
{\displaystyle {\tbinom {m}{r}}}
combinaisons . C'est à dire le nombre de façon de choisir choisir r m m m r m {0,1}, avec r m − r
les Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{4 \choose 2} = 6}
</mo></mrow><mfrac linethickness="0"><mn>
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
(
4
2
)
=
6
{\displaystyle {4 \choose 2}=6}
Pour obtenir de ce modèle le coefficient binomial de Gauss Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr_q}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
(
m
r
)
q
{\displaystyle {\tbinom {m}{r}}_{q}}
q d d Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{4 \choose 2}_q = 1+q+2q^2+q^3+q^4}
</mo></mrow><mfrac linethickness="0"><mn>
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
(
4
2
)
q
=
1
+
q
+
2
q
2
+
q
3
+
q
4
{\displaystyle {4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}}
On peut montrer que les polynômes ainsi définis satisfont les identités de Pascal données ci-dessous et coïncident donc avec les polynômes donnés par les définitions algébriques. Une manière visuelle de comprendre cette définition consiste à associer à chaque mot un chemin traversant une grille rectangulaire avec des côtés de hauteur r m − r
Comme les coefficients binomiaux ordinaires, les coefficients binomaux de Gauss sont symétriques, c'est-à-dire invariants par l'application Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math> r \rightarrow m-r }
</mi><mo stretchy="false">
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
<mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em">
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
{\displaystyle }
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
(
m
r
)
q
=
(
m
m
−
r
)
q
.
{\displaystyle {m \choose r}_{q}={m \choose m-r}_{q}.}
En particulier,
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose 0}_q ={m \choose m}_q=1 \, ,}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
(
m
0
)
q
=
(
m
m
)
q
=
1
,
{\displaystyle {m \choose 0}_{q}={m \choose m}_{q}=1\,,}
<mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em">
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
{\displaystyle }
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
(
m
1
)
q
=
(
m
m
−
1
)
q
=
1
−
q
m
1
−
q
=
1
+
q
+
⋯
+
q
m
−
1
m
≥
1
.
{\displaystyle {m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.}
Le nom coefficient binomial de Gauss vient du fait [réf. nécessaire] q = 1
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><munder><mo movablelimits="true"> <math>\lim_{q \to 1} {m \choose r}_q = {m \choose r}}
</mo><mrow class="MJX-TeXAtom-ORD"><mi>
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
lim
q
→
1
(
m
r
)
q
=
(
m
r
)
{\displaystyle \lim _{q\to 1}{m \choose r}_{q}={m \choose r}}
pour tous m et r .
Les analogues des identités Pascal pour les coefficients binomiaux de Gauss sont
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = q^r {m-1 \choose r}_q + {m-1 \choose r-1}_q}
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
(
m
r
)
q
=
q
r
(
m
−
1
r
)
q
+
(
m
−
1
r
−
1
)
q
{\displaystyle {m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}}
et
<mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em">
(
m
r
)
q
=
(
m
−
1
r
)
q
+
q
m
−
r
(
m
−
1
r
−
1
)
q
.
{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.}
(
m
r
)
q
=
(
m
−
1
r
)
q
+
q
m
−
r
(
m
−
1
r
−
1
)
q
.
{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.}
(
m
r
)
q
=
(
m
−
1
r
)
q
+
q
m
−
r
(
m
−
1
r
−
1
)
q
.
{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.}
(
m
r
)
q
=
(
m
−
1
r
)
q
+
q
m
−
r
(
m
−
1
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{\displaystyle }
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Il existe des analogues de la formule du binôme de Newton et de la version généralisée de celle-ci pour les exposants entiers négatifs, bien que, dans le premier cas, les coefficients binomiaux de gauss eux-mêmes n'apparaissent pas sous forme de coefficients:
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><munderover><mo> <math>\prod_{k=0}^{n-1} (1+q^kt)=\sum_{k=0}^n q^{k(k-1)/2} {n \choose k}_q t^k }
</mo><mrow class="MJX-TeXAtom-ORD"><mi>
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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=
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1
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1
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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=
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1
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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1
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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1
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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k
=
0
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−
1
(
1
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k
t
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
∏
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
∏
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
∏
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
∏
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
∏
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{\displaystyle \prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}}
et
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><munderover><mo> <math>\prod_{k=0}^{\infty} (1+q^kt)=\sum_{k=0}^\infty \frac{q^{k(k-1)/2}t^k}{[k]_q!\,(1-q)^k} }
</mo><mrow class="MJX-TeXAtom-ORD"><mi>
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
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=
∑
k
=
0
∞
q
k
(
k
−
1
)
/
2
t
k
[
k
]
q
!
(
1
−
q
)
k
{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
∏
k
=
0
∞
(
1
+
q
k
t
)
=
∑
k
=
0
∞
q
k
(
k
−
1
)
/
2
t
k
[
k
]
q
!
(
1
−
q
)
k
{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
∏
k
=
0
∞
(
1
+
q
k
t
)
=
∑
k
=
0
∞
q
k
(
k
−
1
)
/
2
t
k
[
k
]
q
!
(
1
−
q
)
k
{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
∏
k
=
0
∞
(
1
+
q
k
t
)
=
∑
k
=
0
∞
q
k
(
k
−
1
)
/
2
t
k
[
k
]
q
!
(
1
−
q
)
k
{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
∏
k
=
0
∞
(
1
+
q
k
t
)
=
∑
k
=
0
∞
q
k
(
k
−
1
)
/
2
t
k
[
k
]
q
!
(
1
−
q
)
k
{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}
qui, pour Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math>n\rightarrow\infty}
</mi><mo stretchy="false">
n
→
∞
{\displaystyle n\rightarrow \infty }
n
→
∞
{\displaystyle n\rightarrow \infty }
n
→
∞
{\displaystyle n\rightarrow \infty }
n
→
∞
{\displaystyle n\rightarrow \infty }
et
La première identité de Pascal permet de calculer les coefficients binomiaux gaussiens de manière récursive (par rapport à m ) en utilisant les valeurs initiales "frontière"
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose m}_q ={m \choose 0}_q=1 }
</mo></mrow><mfrac linethickness="0"><mi>
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
(
m
m
)
q
=
(
m
0
)
q
=
1
{\displaystyle {m \choose m}_{q}={m \choose 0}_{q}=1}
et montre également que les coefficients binomiaux de Gauss sont bien des polynômes (en q ). La seconde identité Pascal découle de la première en faisant le changement Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math> r \rightarrow m-r }
</mi><mo stretchy="false">
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math> r \rightarrow m-r }
</mi><mo stretchy="false">
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
r
→
m
−
r
{\displaystyle r\rightarrow m-r}
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = {{1-q^{m}}\over {1-q^{m-r}}} {m-1 \choose r}_q }
</mo></mrow><mfrac linethickness="0"><mi>
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
(
m
r
)
q
=
1
−
q
m
1
−
q
m
−
r
(
m
−
1
r
)
q
{\displaystyle {m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}}
ce qui conduit (lorsqu'il est appliqué de manière itérative pour m , m - 1, m - 2, ...) à une expression du coefficient binomial de Gauss, telle que donnée dans la définition ci-dessus.
Les coefficients binomiaux de Gauss apparaissent dans le dénombrement des polynômes symétriques et dans la théorie des partitions . Le coefficient en q r
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{n+m \choose m}_q}
</mo></mrow><mfrac linethickness="0"><mrow><mi>
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
(
n
+
m
m
)
q
{\displaystyle {n+m \choose m}_{q}}
est le nombre de partitions de r parmis m ou moins de parties inférieures ou égales à n . De manière équivalente, il s'agit également du nombre de partitions de r parmis n ou moins de parties inférieures ou égales à m .
Les coefficients binomiaux de gauss jouent également un rôle important dans la théorie du dénombrement des espaces projectifs définis sur un corps fini. En particulier, pour chaque corps fini F q q éléments, le coefficient binomial de gauss
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{n \choose k}_q}
</mo></mrow><mfrac linethickness="0"><mi>
(
n
k
)
q
{\displaystyle {n \choose k}_{q}}
(
n
k
)
q
{\displaystyle {n \choose k}_{q}}
(
n
k
)
q
{\displaystyle {n \choose k}_{q}}
(
n
k
)
q
{\displaystyle {n \choose k}_{q}}
(
n
k
)
q
{\displaystyle {n \choose k}_{q}}
(
n
k
)
q
{\displaystyle {n \choose k}_{q}}
compte le nombre de sous-espaces vectoriels de dimension k d'un espace vectoriel de dimension n sur F q grassmannien ). Lorsqu'il est étendu en tant que polynôme en q , il produit la décomposition bien connue du grassmannien en cellules de Schubert. Par exemple, le coefficient binomial de gauss
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{n \choose 1}_q=1+q+q^2+\cdots+q^{n-1}}
</mo></mrow><mfrac linethickness="0"><mi>
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
(
n
1
)
q
=
1
+
q
+
q
2
+
⋯
+
q
n
−
1
{\displaystyle {n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}}
est le nombre de sous-espaces de dimension 1 dans ( F q n espace projectif associé). De plus, lorsque q vaut 1 (respectivement -1), le coefficient binomial de gauss donne la caractéristique d'Euler du complexe de Grassmann complexe (respectivement réel) correspondant.
Le nombre de sous-espaces affines de dimension k de F q n
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msup><mi> <math>q^{n-k} {n \choose k}_q}
</mi><mrow class="MJX-TeXAtom-ORD"><mi>
q
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
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{\displaystyle q^{n-k}{n \choose k}_{q}}
Cela permet une autre interprétation de l'identité
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = {m-1 \choose r}_q + q^{m-r}{m-1 \choose r-1}_q}
</mo></mrow><mfrac linethickness="0"><mi>
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
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{\displaystyle {m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}}
en comptant les sous - espaces de dimension (r - 1) de l'espace projectif de dimension (m - 1), en fixant un hyperplan, en comptant les sous - espaces contenus dans ce hyperplan, puis en comptant les sous - espaces qui ne figurent pas dans l'hyperplan; ces derniers sous-espaces sont en correspondance bijective avec les sous-espaces affines de dimensions ( r - 1) de l'espace obtenu en traitant cet hyperplan fixe comme l'hyperplan à l'infini.
Dans les conventions courantes dans les applications aux groupes quantiques , une définition légèrement différente est utilisée; le coefficient binomial quantique il y a
Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msup><mi> <math>q^{k^2 - n k}{n \choose k}_{q^2}}
</mi><mrow class="MJX-TeXAtom-ORD"><msup><mi>
q
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
q
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{\displaystyle q^{k^{2}-nk}{n \choose k}_{q^{2}}}
Cette version du coefficient binomial quantique est symétrique sous le changement de Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math>q}
</mi></mstyle></mrow> </math>
q
{\displaystyle q}
q
{\displaystyle q}
Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msup><mi> <math>q^{-1}}
</mi><mrow class="MJX-TeXAtom-ORD"><mo>
q
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{\displaystyle q^{-1}}
q
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{\displaystyle q^{-1}}
q
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{\displaystyle q^{-1}}
q
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{\displaystyle q^{-1}}
Les coefficients binomiaux de gauss peuvent être disposés en triangle pour chaque q , pour q=1, on retrouve le triangle de Pascal . </br> Lisez ligne par ligne ces triangles forment les séquences suivantes dans OEIS :
![{\displaystyle [k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d655ed82c2d89a0d675546ead03ce167b7a22e5)
![{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b99b1b84c135ba57a79828e7007620ebf60a09b0)
![{\displaystyle [n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7195538d2b0d91b39b3c2504733378537f18ae59)
![{\displaystyle {m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f186d04b91400b7da6ff2201abe1a83f2bede1e)
![{\displaystyle {\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9f102bb98e4b9cc4f7e5fae7a49e3aa78c0914a)
![{\displaystyle \prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/994e7a8dd56b63ff627a65a3352baa089c6d5121)
