<s>[INST] 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.
 [/INST] 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
$$[z^n] \frac{z^{n-m+r}}{(1-z)^{r-n+1}}
= [z^{n-(n-m+r)}] \frac{1}{(1-z)^{r-n+1}}
\\ = [z^{m-r}] \frac{1}{(1-z)^{r-n+1}} =
{m-r+r-n\choose r-n} = {m-n\choose r-n}.$$
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