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 &gt; m$. Could it be somehow calculated into a shorter an nicer expression which doesn't contain the sum?

Thanks in advance!
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}&amp;\binom{m+a}{k}\binom{m}{k}\\
&amp;=\sum_{k=0}^{\infty}[x^k](1+x)^{m+a}[y^k](1+y)^m\tag{1}\\
&amp;=[x^0](1+x)^{m+a}\sum_{k=0}^{\infty}x^{-k}[y^k](1+y)^m\tag{2}\\
&amp;=[x^0](1+x)^{m+a}(1+\frac{1}{x})^m\tag{3}\\
&amp;=[x^0](1+x)^{m+a}\frac{1}{x^m}(1+x)^m\\
&amp;=[x^m](1+x)^{2m+a}\\
&amp;=\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*}
it is related to a question I answered yesterday :

$$\sum_{k=0}^m {n \choose k} {m \choose k} = \frac1{2\pi}\int_0^{2\pi} (e^{ix}+1)^n (e^{-ix}+1)^m dx =  \frac{2^{n+m}}{2\pi} \int_0^{2\pi}e^{ix(n-m)/2}\cos^{m+n}(x/2) dx $$