TITLE: Evaluating $\sum^n_{k=0}{n\choose{k}}{m\choose{k}}$ for fixed $m,n$
QUESTION [5 upvotes]: For evaluating $\sum^n_{k=0}{n\choose{k}}{m\choose{k}}$ for fixed $m,n$, I got ${n+m\choose{n}}$, does it look right?
I rewrite it as $\sum^n_{k=0}{n\choose{n-k}}{m\choose{k}}$, that's what I did.

REPLY [2 votes]: I think you can have combinatorial arguments for it if you rewrite the expression as $$ \sum_{k=0}^n  \begin{pmatrix}
        n\\
        n-k
       \end{pmatrix} \begin{pmatrix}
        m\\
        k
       \end{pmatrix} $$
This is just equal to choosing $n$ objects out of $n + m $ of them, thus making it equal to $  \begin{pmatrix}
        n+m\\
        n
       \end{pmatrix}$, first keep a particular choice of $m$ objects aside from the rest $n $ objects, then choose some ( say $k$ of them ) from the set of $m$ objects and rest from the set of $n$ objects. This you can do for all $ 0\leq k\leq n $, thus you obtain the expression.