<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\langchapter{Combinatoire}{Combinatorics}
%TODO Complete chapter

\langsection{Formules}{Formulas}

$\prod\limits_{k=1}^{n} k = 1 \times 2 \times 3 \times \cdots \times n = n!$

$\prod\limits_{k=1}^{n} 2k = 2 \times 4 \times 6 \times \cdots \times 2n = 2^n n!$

$\prod\limits_{k=1}^{n} (2k - 1) = 1 \times 3 \times 5 \times \cdots \times (2n + 1) = \frac{(2n + 1)!}{2^n n!}$

$\sum\limits_{k=0}^n \binom{n}{k} = 2^n$

$\binom{n}{k}=\left\{\begin{aligned} &amp;\frac{n!}{k!(n - k)!} &amp; &amp; \text{si } k \in \discreteInterval{0,n} \\ &amp;0 &amp; &amp;\text{sinon} \end{aligned}\right.$

$\forall n \in \N,\forall k \in \Z, \binom{n}{n-k} = \binom{n}{k}$

Formule de Pascal

$\forall n \in \N, \forall k \in \Z, \binom{n}{k - 1} + \binom{n}{k} = \binom{n + 1}{k}$
</pre></body></html>