Blog Archives

A Double Sum of Binomials

Prove the following summation of binomial coefficients. \[ \sum_{i=0}^n \sum_{j=0}^m {i+j \choose j} = {n + m +2 \choose n+1} - 1. \] First, by Pascal’s rule that \({i+j \choose j} = {i+j-1 \choose j} + {i+j-1 \choose j-1} \),

Tagged with:
Posted in Combinatorics