<s>[INST] 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?
 [/INST] 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
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}} 
\sum_{k=0}^n {n\choose k} \frac{(1+w)^k}{z^k}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}} 
\left(1+\frac{1+w}{z}\right)^n
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}} 
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n-m+1}} 
(1+w+z)^n
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^{n+1}} 
{n\choose n-m} (1+z)^{n-(n-m)}
\; dz
\\ = {n\choose m} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+m}}{z^{n+1}} \; dz
\\ = {n\choose m} {n+m\choose n}.$$
</s>