范德蒙德卷积

范德蒙德卷积

\(\displaystyle C _{x+y}^{k}= \sum _{i=0}^{k}C_{x}^{i} \times C_{y}^{k-i}\);

证明：过菜已隐藏。

意会：在\(x+y\)中选k个，相当于在\(x\)和\(y\)中分别选\(i\)个和\(k-i\)个；

所以…… \(C _{x+y}^{k}= \sum _{i=0}^{k}C_{x}^{a+i} \times C_{y}^{k-a-i}\)?

\(eg:\)

求 \(\displaystyle \sum _{i=0}^{x} C_{x}^{i} C_{y}^{z+i} ~mod~998244353\); \((0<=x,y,z<=1000000,且保证y>=z)\)

\(\displaystyle 原式=\sum _{i=0}^{x}C_{x}^{x-i}C_{y}^{z+i}=C_{x+y}^{x+z}\)

\(ex:\)

1. \(\displaystyle C_{2n}^{n-1}= \sum _{i=0}^{n-1}C_{n}^{i}C_{n}^{i-1}= \sum _{i=1}^{n}C_{n}^{i}C_{n}^{i+1}\)

   证：

   \(\displaystyle C_{2n}^{n}= \sum _{i=0}^{n-1}C_{n}^{i}C_{n}^{n-(n-1-i)}= \sum _{i=0}^{n-1}C_{n}^{i}C_{n}^{i+1}= \sum _{i=1}^{n} C_{n}^{i}C_{n}^{i-1}\)

   毕；

2. \(\displaystyle C_{2n}^{n}= \sum _{i=0}^{n}(C_{n}^{i})^2\)

3. \(C_{n+m}^{m}= \sum _{i=0}^{m}C_{n}^{i}C_{m}^{i}\)