帕斯卡法则

${\displaystyle {n \choose k}}$ 表示在有${\displaystyle n}$ 个元素的集内，有${\displaystyle k}$ 个元素的子集的数目。其实这些子集之中，可分为包含第一个元素的和不含第一个元素的。包含第一个元素的子集有${\displaystyle {n-1 \choose k-1}}$ 个，不含的有${\displaystyle {n-1 \choose k}}$ 个。

${\displaystyle {n-1 \choose k}+{n-1 \choose k-1}={n \choose k}}$ 

重写左边为

${\displaystyle {\begin{aligned}&{}{\frac {(n-1)!}{k!(n-k-1)!}}+{\frac {(n-1)!}{(k-1)!(n-k)!}}\\&={\frac {(n-k)(n-1)!}{(n-k-1)!k!(n-k)}}+{\frac {k(n-1)!}{k(k-1)!(n-k)!}}\\&={\frac {(n-k)(n-1)!+k(n-1)!}{k!(n-k)!}}\\&={\frac {(n-1)!\times [(n-k)+k]}{k!(n-k)!}}\\&={\frac {(n-1)!\times n}{k!(n-k)!}}\\&={\frac {n!}{k!(n-k)!}}\\&={n \choose k}\end{aligned}}}$ 

设${\displaystyle n,k_{1},k_{2},k_{3},\dots ,k_{p},p\in \mathbb {N} ^{*}\,\!}$ 及${\displaystyle n=k_{1}+k_{2}+k_{3}+\cdots +k_{p}\,\!}$ 。那么：

${\displaystyle {\begin{aligned}&{}\quad {n-1 \choose k_{1}-1,k_{2},k_{3},\dots ,k_{p}}+{n-1 \choose k_{1},k_{2}-1,k_{3},\dots ,k_{p}}+\cdots +{n-1 \choose k_{1},k_{2},k_{3},\dots ,k_{p}-1}\\&={\frac {(n-1)!}{(k_{1}-1)!k_{2}!k_{3}!\cdots k_{p}!}}+{\frac {(n-1)!}{k_{1}!(k_{2}-1)!k_{3}!\cdots k_{p}!}}+\cdots +{\frac {(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots (k_{p}-1)!}}\\&={\frac {k_{1}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}+{\frac {k_{2}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}+\cdots +{\frac {k_{p}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}\\&={\frac {(k_{1}+k_{2}+\cdots +k_{p})(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}\\&={\frac {n(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}\\&={\frac {n!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}\\&={n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}\end{aligned}}}$ 

• 杨辉三角形