$\sum_{i=0}^{k} \binom{m}{i}\binom{n}{k-i} =\binom{m+n}{k}$

Question

I'm trying to show that the equality $$\sum_{i=0}^{k} \binom{m}{i}\binom{n}{k-i} =\binom{m+n}{k}$$

Is true. I know it is since there is a good combinatorical argument for it. If we have a group of $m$ men and $n$ women, and we need to choose $k$ then we know its equal to $\binom{m+n}{k}$, but we can also say that we choose $0$ men and $k$ women, or we can choose $1$ man and $k-1$ women, or we can choose $2$ men and $k-2$ women, etc.

So in theory they should be equal, but I can't work the math behind it.

A hint in the right direction would be appreciated

'So in theory they should be equal, but I can't work the math behind it.' .. not sure what you are asking. — sciona, Jan 18 '15 at 21:56
I'm asking for a formal mathematical proof for this equality. Not a combinatorical proof where we count the same thing in 2 ways like i did. — Oria Gruber, Jan 18 '15 at 21:57
Another combinatorial argument, if one does not like men or women. The left-hand side is the coefficient of $x^k$ in $(1+x)^m(1+x)^n$. — André Nicolas, Jan 18 '15 at 21:58

score 3 · Accepted Answer · answered Jan 18 '15 at 22:04

Consider the identity $$(1+x)^m(1+x)^n=(1+x)^{m+n}$$ Develop the LHS using the binomial identity for $(1+x)^m$ and $(1+x)^n$ and do the same with the RHS using the binomial identity for $(1+x)^{m+n}$ and then identify the coefficients of the term of degree $k$ on each side. It is exactly what you are trying to prove.

$\sum_{i=0}^{k} \binom{m}{i}\binom{n}{k-i} =\binom{m+n}{k}$

1 Answers1