Is there any way to calculate the simple product of binomial coefficients Given the sum $$ \sum_{k=0}^{m} {n \choose k} {m \choose k}, $$ where $ n > m$. Could it be somehow calculated into a shorter an nicer expression which doesn't contain the sum?
Thanks in advance!
 A: We use the  coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ of a series. So, we can write e.g.
 $$\binom{m}{k}=[x^k](1+x)^m$$

We obtain for $a\geq 0$
  \begin{align*}
\sum_{k=0}^{m}&\binom{m+a}{k}\binom{m}{k}\\
&=\sum_{k=0}^{\infty}[x^k](1+x)^{m+a}[y^k](1+y)^m\tag{1}\\
&=[x^0](1+x)^{m+a}\sum_{k=0}^{\infty}x^{-k}[y^k](1+y)^m\tag{2}\\
&=[x^0](1+x)^{m+a}(1+\frac{1}{x})^m\tag{3}\\
&=[x^0](1+x)^{m+a}\frac{1}{x^m}(1+x)^m\\
&=[x^m](1+x)^{2m+a}\\
&=\binom{2m+a}{m}
\end{align*}

Comment:


*

*In (1) we use the coefficient of Operator and change the limit to $\infty$ without changing the sum, since we add only zero.

*In (2) we use the linearity of the coefficient of operator and $[x^{n+k}]A(x)=[x^n]x^{-k}A(x)$

*In (3) we use the substitution rule
\begin{align*}
A(x)=\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{\infty}x^k[y^k]A(y)
\end{align*}
A: Note that
$$\sum_{k=0}^m\binom{n}k\binom{m}k=\sum_{k=0}^m\binom{n}k\binom{m}{m-k}\;.\tag{1}$$
Suppose that we have a pool of $n$ women and $m$ men, from which we are to choose a committee of $m$ people; clearly the righthand side of $(1)$ is the number possible committees, counted according to the number ($k$) of women on the committee.
On the other hand, we can simply choose any $m$ of the $n+m$ members of the pool. This can be done in $\binom{n+m}m$ ways, so
$$\sum_{k=0}^m\binom{n}k\binom{m}k=\sum_{k=0}^m\binom{n}k\binom{m}{m-k}=\binom{n+m}m\;.$$
This is an instance of Vandermonde’s identity, and the argument that I just gave is the usual combinatorial proof of it.
A: it is related to a question I answered yesterday :
$$\sum_{k=0}^m {n \choose k} {m \choose k} = \frac1{2\pi}\int_0^{2\pi} (e^{ix}+1)^n (e^{-ix}+1)^m dx =  \frac{2^{n+m}}{2\pi} \int_0^{2\pi}e^{ix(n-m)/2}\cos^{m+n}(x/2) dx $$
