Combinatorial Proof of $\binom{n+m}{k} = \sum_{i=0}^k \binom{n}{i} \binom{m}{k-i}$ I want to prove the following combinatorially:
$$\binom{n+m}{k} = \sum_{i=0}^k \binom{n}{i} \binom{m}{k-i}$$
I know how to solve this algebraically but I need a solution based on ideas of combinatorics (like a background story and such). 
Any help would be appreciated.
 A: The property you just asked 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: You have $n$ girls and $m$ boys. Therefore, you have $m+n$ people. How many possibilities to make a group with $k$ people ? To make such a group you can also make groups with only boys, or 1 girl and $k-1$ boys, or $2$ girls and $k-2$ boys... At the end you get the wished result.
A: Paint $n$ of the original $n+m$ objects blue and the other $m$ red. How many ways can we choose $k$ objects? The left-hand side is the correct answer; the right-hand side writes it as the sum of all ways of doing it with a specified number of blue objects.
A: Not exactly combinatorial, but there are already three combinatorial answers.
What is the coefficient of $x^ky^{m+n-k}$ in $(x+y)^{m+n}$? And in $(x+y)^m(x+y)^n$?
