Read · 1 minBy AAhaan RungtaVandermonde's Identity Let m,n,r∈N0m,n,r \in \mathbb{N}_0m,n,r∈N0. Then: (m+nr)=∑k=0r(mk)(nr−k).\dbinom {m+n}{r} = \displaystyle\sum_{k=0}^{r} \dbinom {m}{k} \dbinom {n}{r-k}.(rm+n)=k=0∑r(km)(r−kn). ReportReport a problemWhat needs fixing?Send reportShareLike 2Share explanationExplanation linkCopy link