Exponential generating function problem involving product formula For any integer $n\geq0$, let 
$f_n=$ $\sum_{k=0}^∞$${n}\choose{k}$${k}\choose{r}$$(-1)^{n-k}$${n-k}\choose{r}$
How do I show that $f_{2r}=(-1)^{r}$${2r}\choose{r}$ and $f_n=0$ if $n\ne2r$?
The only hint I was given was to apply the product formula to calculate the exponential generating function of the sequence $\{f_n\}$, but I am also struggling with this.
Any help would be greatly appreciated, thank you. 
 A: I think this  is what the question is looking  for.  Observe that when
we  multiply two  exponential  generating functions  of the  sequences
$\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
Therefore with
$$F(z) = \sum_{n\ge 0} \frac{z^n}{n!}
\sum_{k=0}^n {n\choose k} {k\choose r} (-1)^{n-k} {n-k\choose r}$$
we have $$F(z) = A(z) B(z)$$ where
$$A(z) = \sum_{n\ge 0} {n\choose r} \frac{z^n}{n!}
\quad\text{and}\quad
B(z) = \sum_{n\ge 0} {n\choose r} (-1)^r \frac{z^n}{n!}.$$
We get for $A(z)$
$$A(z) = \sum_{n\ge r} {n\choose r} \frac{z^n}{n!}
= \frac{1}{r!} \sum_{n\ge r} \frac{1}{(n-r)!} z^n
\\ = \frac{z^r}{r!} \sum_{n\ge 0} \frac{1}{n!} z^n
= \frac{z^r}{r!} \exp(z).$$
and for $B(z)$
$$B(z) = \sum_{n\ge r} {n\choose r} (-1)^n \frac{z^n}{n!}
= \frac{(-1)^r}{r!} \sum_{n\ge r} \frac{(-1)^{n-r}}{(n-r)!} z^n
\\ = (-1)^r \frac{z^r}{r!} \sum_{n\ge 0} \frac{(-1)^n}{n!} z^n
= (-1)^r \frac{z^r}{r!} \exp(-z).$$
It follows that
$$F(z) = \frac{z^r}{r!} \exp(z) (-1)^r \frac{z^r}{r!} \exp(-z)
= (-1)^r \frac{z^{2r}}{r!\times r!}.$$
Therefore
$$f_n = n! [z^n] F(z) = n! [z^n] (-1)^r \frac{z^{2r}}{r!\times r!}
\\ = [[n=2r]] n! (-1)^r \frac{1}{r!\times r!}
= [[n=2r]] (-1)^r \times {2r\choose r}.$$
This is the claim.
A: Using the coefficient extractor thingy, Eg.
\begin{eqnarray*}
\binom{k}{r} = [x^r]: (1+x)^k.
\end{eqnarray*}
We have
\begin{eqnarray*}
f_n &=& \sum_{k} (-1)^{n-k} \binom{n}{k}  \binom{k}{r} \binom{n-k}{r} \\  
&=& [x^r] [ y^r] :  \sum_{k} (-1)^{n-k} \binom{n}{k}(1+x)^{k} (1+y)^{n-k}  \\  
&=& [x^r] [ y^r] :  (-1)^{n}  (1+y)^{n} \left( 1-\frac{1+x}{1+y} \right)^n \\  
&=& [x^r] [ y^r] :  (-1)^{n}  (y-x)^{n} \\  
\end{eqnarray*}
And the only way to get a $x^ry^r$ term is if $n=2r$, the result follows.
