Prove the following identity : a)$$ \binom{n+1}{2m+1}=\sum_{k=0}^n \binom{n-k}{m} \cdot \binom{k}{m} $$ where $n$ and $m$ are positive integers
$\binom{n}{m}=\frac{n!}{(n-m)!m!}$ , for $n\geq m$
and $\binom{n}{m}=0$, for $n<m$
Idea: If we choose $2m+1$ numbers from $n+1$ then one number is in the middle. If this number is $k+1$ then $0\le k\le n$ and we have to choose $m$ numbers to the left to it, ${k\choose m}$ and $m$ numbers to the right of it, ${{n-k}\choose m}$.
How to proceed with this idea?
