TITLE: Combinatorial interpretation for the identity $\sum\limits_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}$?
QUESTION [5 upvotes]: A known identity of binomial coefficients is that
$$
\sum_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}.
$$
Is there a combinatorial proof/explanation of why it holds? Thanks.

REPLY [2 votes]: What you need to prove is the Vandermonde's identity. 
Vandermonde's Identity states that $\sum_{k=0}^r\binom mk\binom n{r-k}=\binom{m+n}r$, which can be proven combinatorially by noting that any combination of $r$ objects from a group of $m+n$ objects must have some $0\le k\le r$ objects from group $m$ and the remaining from group $n$.