TITLE: Binomial series with two binomial coefficents
QUESTION [5 upvotes]: My question reads: Does this formula has mathematical meaning at first place? Is it summable?
$$\sum^{\infty}_{k=0}{n\choose k}{m\choose k} x^k$$

REPLY [6 votes]: It's quite easy to verify that your sum is equal to the coefficient of $z^m$ in the product:
$$ (1+xz)^n\,(1+z)^m.$$
If you set $x=1$ you can find a slight generalization of the Chu-Vandermonde identity:
$$ \sum_{k=0}^{+\infty}\binom{m}{k}\binom{n}{k}=\binom{m+n}{n}.$$