# Combinatorial argument for identity with binomial coefficients

My goal is to show :

$$\sum_{i=0}^{m} {n \choose i} {n - i \choose m - i} = {n \choose m} 2^m$$ where $$0 \leq m \leq n$$ I have algebraically proved that but I was wondering if there is a combinatorial argument for this proof. Thank you

• Your equation makes no sense: $i$ is a dummy variable with no fixed value on the lefthand side and a fixed constant of some kind on the righthand side. Commented Oct 4, 2020 at 16:51
• That $i$ on RHS should probably be $m$. Commented Oct 4, 2020 at 16:52
• I am sorry there was a typo. I fixed it Commented Oct 4, 2020 at 16:53

There are two ways to count the number of subsets $$A,B\subset\{1,\ldots,n\}$$ with $$A\subset B$$ and $$|B|=m$$:
On the left you pick any subset $$A$$ of at most $$m$$ elements, and then extend it to a subset $$B$$ of $$m$$ elements. On the right you pick a subset $$B$$ of $$m$$ elements, and then pick any subset $$A$$ of $$B$$.
• Very neat!${}{}$ Commented Oct 4, 2020 at 16:57