$\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}$ using Counting argument

I saw this question here:- Combinatorial sum identity for a choose function

This looks so much like a vandermonde identity, I know we can give a counting argument for Vandermonde. However much I try I am not able to come up with a counting argument for this.

Let $S=\{-m,-m+1,\ldots,n-1,n\}$; $|S|=m+n+1$, and we want to count the subsets of $S$ of cardinality $r+s+1$. Suppose that $A$ is such a subset. Then there is a unique $k_A\in A$ such that $r$ members of $A$ are smaller than $k_A$, and $s$ members of $A$ are larger than $k_A$. Let $\mathscr{A}_k$ be the family of $(r+s+1)$-element subsets $A$ of $S$ such that $k_A=k$. There are $k-(-m)=m+k$ elements of $S$ less than $k$ and $n-k$ elements of $S$ greater than $k$, so

$$|\mathscr{A}_k|=\binom{m+k}r\binom{n-k}s\;.$$

Summing over $k$ gives us the total number of $(r+s+1)$-element subsets of $S$, which is of course $\binom{m+n+1}{r+s+1}$, so

$$\sum_{k=-m}^n\binom{m+k}r\binom{n-k}s=\binom{m+n+1}{r+s+1}\;.$$

You have $m+n+2$ identical coins, and $r+s+2$ distinct boxes.
How many ways are there to place all the coins in the boxes so no box is empty?

If you place these $m+n+2$ coins in a line, then there are $m+n+1$ spaces between pairs of adjacent coins, and we can choose $r+s+1$ of these spaces to place a divider in. These dividers split the row of coins into $r+s+2$ contiguous blocks, representing the $r+s+2$ boxes, and therefore uniquely describing a placement of coins in boxes. Therefore, the number of ways to place the coins is $\binom{m+n+2}{r+s+1}$.

On the other hand, we can instead count how many ways to place the coins there are where the first $r+1$ boxes receive $m+k+1$ coins total, meaning the remaining $s+1$ boxes receive the remaining $n-k+1$ coins, and then sum over $k$. By the same argument, there are $\binom{m+k}r$ ways to place the $m+k+1$ coins in the first $r+1$ boxes, and $\binom{n-k}{s}$ ways to place the remaining $n-k+1$ coins in the other $s+1$ boxes.