How to prove $\sum_{k=0}^n {n\choose k}^2 {k\choose {n-m}} = {n\choose m} {{m+n}\choose m}$? $$\sum_{k=0}^n {n\choose k}^2 {k\choose {n-m}} = {n\choose m} {{m+n}\choose m}$$
I am struggling with this identity, with no progress. Can someone show me the proof?
 A: Suppose that you have $n$ women in Room A and $n$ men in Room B. You choose $m$ of the women and move them into Room B, and then you choose $m$ of the people in Room B to move into Room C. When you’re done with this, you have $n-m$ women in Room A, $n$ people in Room B, and $m$ people in Room C, and the total number of men in Rooms B and C is $n$. There are clearly $\binom{n}m\binom{m+n}m$ possible outcomes.
Now count according to the number $k$ of men who end up in Room B, which may be anywhere from $0$ to $n$ inclusive. If $k$ men end up in Room B, there must be $n-k$ women who end up in Room B, so there must be $k$ women in Rooms A and C combined. There are $\binom{n}k$ ways to choose the men who end up in Room B, and independently there are $\binom{n}k$ ways to choose the women who end up in Rooms A and C combined. Can you finish it from here? If not, mouse-over the spoiler-protected block below for a final hint.

 All that remains is to choose which of the women in Rooms A and C combined are actually to end up in Room A.

A: Suppose we seek to verify that
$$\sum_{k=0}^{n}  {n\choose k}^2 {k\choose n-m}
= {n\choose m} {m+n\choose m}.$$
where presumably $n\ge m.$
Introduce
$${n\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{k+1}} \; dz.$$
and
$${k\choose n-m}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{k}}{w^{n-m+1}} \; dw.$$
This gives for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}} 
\sum_{k=0}^n {n\choose k} \frac{(1+w)^k}{z^k}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}} 
\left(1+\frac{1+w}{z}\right)^n
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}} 
(1+w+z)^n
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}} 
{n\choose n-m} (1+z)^{n-(n-m)}
\; dz
\\ = {n\choose m} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+m}}{z^{n+1}} \; dz
\\ = {n\choose m} {n+m\choose n}.$$
