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

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

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