Proving $\sum\limits_{k=0}^{\infty}\binom {m-r+s}k\binom {n+r-s}{n-k}\binom {r+k}{m+n}=\binom rm\binom sn$ 
Question: How do you show the following equality holds using binomials$$\sum\limits_{k=0}^{\infty}\binom {m-r+s}k\binom {r+k}{m+n}\binom {n+r-s}{n-k}=\binom rm\binom sn$$

I would like to prove the identity using some sort of binomial identity. The right-hand side is the coefficient of $x^m$ and $y^n$ in$$\begin{align*}a_{m,n} & =\left[x^m\right]\left[y^n\right](1+x)^r(1+y)^s\\ & =\binom rm\binom sn\end{align*}$$
However, I don’t see how the left-hand side can be proven using the binomials. Using the generalized binomial theorem, we get the right-hand side as
$$\begin{align*}(1+x)^r(1+y)^s & =\sum\limits_{k\geq0}\sum\limits_{l\geq0}\binom rk\binom slx^ky^l\end{align*}$$However, what do I do from here?

With OP asking for formal power series in the evaluation of
$$\sum_{k\ge 0} {m-r+s\choose k} {r+k\choose m+n}
{n+r-s\choose n-k}$$
we write
$$[z^n] (1+z)^{n+r-s}  [w^{m+n}] (1+w)^r
\sum_{k\ge 0} {m-r+s\choose k}
z^k (1+w)^k
\\ = [z^n] (1+z)^{n+r-s}  [w^{m+n}] (1+w)^r
(1+z+zw)^{m-r+s}
\\ = [z^n] (1+z)^{n+r-s}  [w^{m+n}] (1+w)^r
\sum_{q=0}^{m-r+s} {m-r+s\choose q}
(1+z)^{m-r+s-q} z^q w^q
\\ = \sum_{q=0}^{m-r+s} {m-r+s\choose q}
{m+n-q\choose n-q} {r\choose m+n-q}.$$
Note that
$${m+n-q\choose n-q} {r\choose m+n-q}
= \frac{r!}{(n-q)! \times m! \times (r+q-m-n)!}
\\ = {r\choose m} {r-m\choose n-q}.$$
We thus have
$${r\choose m}
\sum_{q=0}^{m-r+s} {m-r+s\choose q} {r-m\choose n-q}
\\ = {r\choose m} [z^n] (1+z)^{r-m}
\sum_{q=0}^{m-r+s} {m-r+s\choose q} z^q
\\ = {r\choose m} [z^n] (1+z)^{r-m} (1+z)^{m-r+s}
= {r\choose m} {s\choose n}.$$
This is the claim.
