$ \large1.$ 容斥原理
\[f(n) = \sum_{i = 0}^n \dbinom{n}{i} g(i) \Leftrightarrow g(n) = \sum_{i = 0}^n (-1)^{n-i} \dbinom{n}{i}f(i) \]$\large f $ 表示至多 ,\(\large g\) 表示恰好
\[f(n) = \sum_{i = n}^m \dbinom{i}{n} g(i) \Leftrightarrow g(n) = \sum_{i = n}^m (-1)^{i-n} \dbinom{i}{n}f(i) \] \(\large f\) 表示至少 , $ \large g $ 表示恰好
\(\large2.\) 组合恒等式
\[\dbinom{r}{k} = \dbinom{r-1}{k} + \dbinom{r-1}{k-1} \]\[\dbinom{r}{m}\dbinom{m}{k} = \dbinom{r}{k}\dbinom{r-k}{m-k} \]\[\sum_{k} \dbinom{r}{k} x^k y^{r-k} = (x + y)^r \]\[\sum_{k\leq n}\dbinom{n+k}{k} = \dbinom{r+n+1}{n} \]\[\sum_{0\leq k \leq n} \dbinom{k}{m} = \dbinom{n+1}{m+1} \]$ \large 3.$ 范德蒙德卷积
\[\sum_k\dbinom{l}{m+k}\dbinom{s}{n-k} = \dbinom{l+s}{m+n} \]\[\sum_k \dbinom{l}{m+k}\dbinom{s}{n+k} = \dbinom{l+s}{l - m + n} \]\[\sum_k \dbinom{l-k}{m}\dbinom{s+k}{n} = \dbinom{l + s + 1}{m+n+1} \] 标签：dbinom,组合,sum,large,leq,数学,Leftrightarrow From： https://www.cnblogs.com/WG-MingJunYi/p/18307498