组合

组合数公式

• $\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)$
• $\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{i}{m}\right)=\left(\genfrac{}{}{0ex}{}{n}{m}\right)\left(\genfrac{}{}{0ex}{}{n-m}{i-m}\right)$
• $\sum _{r=0}^k\left(\genfrac{}{}{0ex}{}{n+r-1}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+k}{k}\right)$
• $\sum _{i=m}^n\left(\genfrac{}{}{0ex}{}{a+i}{i}\right)=\left(\genfrac{}{}{0ex}{}{a+n+1}{n}\right)-\left(\genfrac{}{}{0ex}{}{a+m}{m-1}\right)$
• $\sum _{i=m}^n\left(\genfrac{}{}{0ex}{}{i}{a}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{a+1}\right)-\left(\genfrac{}{}{0ex}{}{m}{a+1}\right)$

常见模型

1. 从 $n$ 堆元素（每堆有无穷多个）中选出 $k$ 个无序的方案数 设第 $i$ 堆中选择的个数为 $x_i$，则有 ${\sum }_i{x}_i=k$，然后可以用插板法解决。