Evaluate $\sum_{k=0}^n (-1)^k{n \choose k}{m-k \choose r}$ using generating functions. I am trying to evaluate this with generating functions under the assumption that $m \geq n$.  From there I need to come up with a proof of the identity obtained using principle of inclusion exclusion.
I tried to set this up as a binomial convolution but can't seem to make it work.
 A: I don’t at the moment see how to do it with generating functions, but I can point you at an inclusion-exclusion argument that makes it pretty easy to see what the identity is. Perhaps once you know that, you can reverse engineer it to get the generating function argument.
You have a pool of $m$ people; $n$ of them are women, and the rest are men. From this pool you want to form a committee of $r$ people that includes all of the women. The sum is an inclusion-exclusion calculation of the number of ways to form the committee. It’s not hard to find a simple expression for the number of ways to form the committee.
A: Here is the closed form with complex analysis.

Suppose we are interested in
$$\sum_{k=0}^n (-1)^k {n\choose k}
{m-k\choose r}$$
where  $m\ge n$  and $r\ge  n.$ We  will first  solve this  by complex
variables and that will tell us what the generating functions are.
Introduce
$${m-k\choose r} = 
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m-k} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m} 
\sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m} 
\left(1-\frac{1}{1+z}\right)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m-n} 
z^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r-n+1}} (1+z)^{m-n} 
\; dz
\\ = {m-n\choose r-n}.$$
Now  that we  have this  we  can reverse-engineer  the solution  using
generating functions.
We thus get 
$${n\choose k} (-1)^k = 
[z^k]\sum_{k=0}^n (-1)^k {n\choose k} z^k = [z^k] (1-z)^n$$
and $${m-k\choose r} = {m-k-r + r \choose r}
= [z^{m-k-r}] \frac{1}{(1-z)^{r+1}}
\\ = [z^{n+m-k-r}] \frac{z^n}{(1-z)^{r+1}}
\\ = [z^{n-k}] \frac{z^{n-m+r}}{(1-z)^{r+1}}.$$
Therefore the desired OGFs are
$$(1-z)^n \quad\text{and}\quad \frac{z^{n-m+r}}{(1-z)^{r+1}}.$$
Doing the multiplication we get
$$\frac{z^{n-m+r}}{(1-z)^{r-n+1}}.$$
Extract the coefficient on $[z^n]$ we have
$$[z^n] \frac{z^{n-m+r}}{(1-z)^{r-n+1}}
= [z^{n-(n-m+r)}] \frac{1}{(1-z)^{r-n+1}}
\\ = [z^{m-r}] \frac{1}{(1-z)^{r-n+1}} =
{m-r+r-n\choose r-n} = {m-n\choose r-n}.$$
