Combinatorial interpretation for the identity $\sum\limits_i\binom{m}{i}\binom{n}{j-i}=\binom{m+n}{j}$? 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.
 A: 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 $n$.
A: The captain of the pirate starship Free Enterprise finds $m$ Martians and $n$ Neptunians in a bar.  In how many ways can she select a crew of $j$ creatures  for a raid on Jupiter?  
The right-hand side $\binom{m+n}{j}$ counts the number of ways to select $j$ creatures from the $m+n$ creatures available.
So does the left-hand side.  For she could select $0$ Martians and $j$ Neptunians. This can be done in $\binom{m}{0}\binom{n}{j}$ ways.
Or else she could select $1$ Martian and $j-1$ Neptunians. This can be done in $\binom{m}{1}\binom{n}{j-1}$ ways.
Or else she could select $2$ Martians and $j-2$ Neptunians.  This can be done in $\binom{m}{2}\binom{n}{j-2}$ ways.
And so on.  Thus the number of ways to recruit $j$ creatures is given by the sum on the left-hand side.
Comment: The result, and the reasoning, remain correct even if, for example, $j>n$. All we need to do is to define  $\binom{u}{v}$ to be $0$ if $u<v$.
