I need a combinatorial proof for the following expression.
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n}{p+k}=\binom{m+n}{m+p}$$
My attempts:
For the left-hand side I tried to create a scenario where I choose first 0 elements from a set of m-elements and a p elements from a set of n-elements then adding the possibilities of choosing 2 more elements where a one from m-elements and p+1 from a set of n elements and so on...
$$\displaystyle\binom{m}{0}\displaystyle\binom{n}{p+0}+\displaystyle\binom{m}{1}\displaystyle\binom{n}{p+1}+\displaystyle\binom{m}{2}\displaystyle\binom{n}{p+2}...$$
I couldn't apply the same scenario on the right-hand side, I feel I didn't put so much creativity in the scenario , it just really translating the left-hand side literally without any twist or so that could help me applying the same scenario on the right-hand side.
Let there be $m$ men and $n$ women in a committee. We want to form a subcommittee consisting of $m + p$ members. Counting directly, we find that there are $\binom{m + n}{m + p}$ ways to form this subcommittee, which is the RHS of the equation.
Another way to approach this problem is to consider how many subcommittees can be formed if we want to exclude $k$ men from the subcommittee, where $k$ can take on any value from $0$ to $m$. Excluding $k$ men from the committee is the same as choosing $m - k$ men to be on the subcommittee. There are $\binom{m}{k} = \binom{m}{m - k}$ ways to do this. To complete the selection, we must pick $(m + p) - (m - k) = p + k$ women from the $n$ women total. There are $\binom{n}{p + k}$ ways to do this. So, for a given $k$, there are $\binom{m}{k} \binom{n}{p + k}$ ways to form the subcommittee. Since the subcommittees formed from differing values of $k$ are distinct, the total number of ways is obtained simply by summing over all possible values of $k$: $\sum_{k = 0}^m \binom{m}{k} \binom{n}{p + k}$, which is the LHS of the equation.