How to prove this identity using combinatorial argument?
$\displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{p}{r+s}\binom{q+r}{m+n} = \sum_{r=0}^{n}\binom{n}{r}\binom{q}{r+m}\binom{p+r}{n+s}$
Only thing I could think of was to change $\binom{n}{r}$ to $\binom{n}{n-r}$ and then we can probably think of choosing $(n-r+r+s+m+n) = (n+s+m+n)$ people or something but its dead end for me. Please help. Also it seems if I can figure one side, other might trivially follow. Please don't ignore the problem.
Best Answer
Suppose you have a fictional country where there are two big parties. The $\color{red}{red}$ and the $\color{blue}{blue}$ parties. Initially, there are $\color{red}{q}$ and $\color{blue}{p}$ of each party. Suppose you want to pick a political committee (senate?) consisting of $\color{red}{(n+m)}$ and $\color{blue}{(n+s)}$ members. Assume, also, that you have a pool of $n$ candidates that can turn red or blue, they are somewhat neutral and can represent either party. By law you have to choose at least $\color{red}{m}$ from the original red party and at least $\color{blue}{s}$ from the original blue party.
There are two ways this elections will happen. Either first the red party chooses or the blue party chooses.
For the LHS suppose that first the $\color{blue}{blue}$ party chooses. Assume that $r+s$ is the number of people selected from the original party. Then we need to choose $n-r$ from the pool. This we can do in $\binom{p}{r+s}\binom{n}{n-r}$ ways. Then, the rest of the pool ($r$ of them) will become $\color{red}{red},$ so in the other election you have to choose from $q+r$ candidates, giving you $\binom{q+r}{n+m}$ ways. This gives us a total of $\binom{p}{r+s}\binom{n}{n-r}\binom{q+r}{n+m}$ ways to do this election.
The RHS is the same, but the $\color{red}{red}$ people vote first.