组合恒等式

组合恒等式

\[\binom n m = \binom n {n - m} \]

\[\sum_{i = 0}^n \binom n i = 2 ^ n \]

\[\binom n m = \binom {n - 1} {m - 1} + \binom {n - 1}{m} \]


用最后一个调整:

\[\sum_{i = 0} ^ n \binom n i [2 | i] = 2 ^ {n - 1} \]


k 个非负整数和为 n 的方案数(插板法) :

\[\binom {n + k - 1} { k - 1 } \]


\[\sum_{i = 0} ^ m \binom {n + i} n = \binom {n + m + 1} m \]

\[\sum_{i = m} ^ n \binom i m = \binom {n + 1} {m + 1} \]

证明:下面 = 上面


\[\binom n m \binom m k = \binom n k \binom {n - k} {m - k} \]


\[\sum_{i = 0} ^ k \binom n i \binom m {k - i} = \binom {n + m} k \]

posted @ 2020-06-06 12:03  __int256  阅读(293)  评论(0编辑  收藏  举报