Another variant of vandermonde's identity: $\binom{r}{m+k} \binom{s}{n+k}$ It can be shown that $$\sum_{k}\binom{r}{m+k} \binom{s}{n+k} = \binom{r+s}{r-m+n}$$ using either lattice paths or manipulation of the binomial coefficients in the identity. How so? I've played with this for hours, and the only worthwhile offering (which is not that hard to guess) I have is that on the lattice of dimension
$$(r-m+n) \times (s-n+m)$$
which represents the RHS binomial coefficient,
the lattice paths represented by $\binom{r}{m+k}$ lie on it. In particular, their right-most corners lie on one line.
 A: Since $\binom{r}{m+k}=\binom{r}{r-m-k}$, the desired identity can be written
$$\sum_k\binom{r}{r-m-k}\binom{s}{n+k}=\binom{r+s}{r-m+n}\;.\tag{1}$$
Now let $\ell=n+k$, so that $k=\ell-n$, and the lefthand side of $(1)$ becomes
$$\sum_\ell\binom{r}{r-m-(\ell-n)}\binom{s}\ell=\sum_\ell\binom{r}{r-m+n-\ell}\binom{s}\ell\;,$$
which by Vandermonde’s identity is equal to $\binom{r+s}{r-m+n}$.
A: Observe that we certainly get all non-zero values when $k\ge -m-n$
so we may write
$$\sum_k {r\choose m+k} {s\choose n+k}
= \sum_{k\ge -m-n} {r\choose m+k} {s\choose n+k}
\\ = \sum_{k\ge 0} {r\choose k-n} {s\choose k-m}
= \sum_{k\ge 0} {r\choose r+n-k} {s\choose s+m-k}.$$
Now put 
$${r\choose r+n-k} =
\int_{|z|=\epsilon}
\frac{1}{z^{r+n-k+1}} (1+z)^r \; dz$$
and
$${s\choose s+m-k} =
\int_{|w|=\gamma}
\frac{1}{w^{s+m-k+1}} (1+w)^s \; dw$$
so that we obtain for the sum
$$\int_{|z|=\epsilon}
\frac{1}{z^{r+n+1}} (1+z)^r 
\int_{|w|=\gamma}
\frac{1}{w^{s+m+1}} (1+w)^s 
\sum_{k\ge 0} z^k w^k
\; dw\; dz
\\ = \int_{|z|=\epsilon}
\frac{1}{z^{r+n+1}} (1+z)^r 
\int_{|w|=\gamma}
\frac{1}{w^{s+m+1}} (1+w)^s 
\frac{1}{1-wz}
\; dw\; dz.$$
This converges for $|zw|\lt 1.$ Now in the inner integral residues sum
to zero.  We obtain the desired sum  for the residue from  the pole at
$w=0.$ Write for the residue from the pole at $w=1/z$
$$-\int_{|z|=\epsilon}
\frac{1}{z^{r+n+2}} (1+z)^r 
\int_{|w|=\gamma}
\frac{1}{w^{s+m+1}} (1+w)^s 
\frac{1}{w-1/z}
\; dw\; dz$$ 
for a contribution of
$$-\int_{|z|=\epsilon}
\frac{1}{z^{r+n+2}} (1+z)^r 
z^{s+m+1} \frac{(1+z)^s}{z^s} \; dz
= -\int_{|z|=\epsilon}
\frac{1}{z^{r+n-m+1}} (1+z)^{r+s} 
\; dz
\\ = -{r+s\choose r-m+n}.$$
For the residue at infinity we get
$$\mathrm{Res}_{w=\infty} \frac{1}{w^{s+m+1}} (1+w)^s 
\frac{1}{1-wz}
\\ = -\mathrm{Res}_{w=0} \frac{1}{w^2} w^{s+m+1}
\frac{(1+w)^s}{w^s} \frac{1}{1-z/w}
= -\mathrm{Res}_{w=0} w^{m} \frac{(1+w)^s}{w-z} = 0.$$
Hence we have the closed form
$$\bbox[5px,border:2px solid #00A000]{
{r+s\choose r-m+n} = {r+s\choose s-n+m}.}$$
