There are $2n$ voters, each write his name on a paper as the voter and the name of his nominee. How many ways there are such that there are exactly $n$ different nominees and each of the nominees has exactly two votes.
Attempt 1:
$\binom {2n} n \sum _{\text{k is even}}^ n \binom {2n-k} 2$
Choose the nominees and then each time choose two voters.
But this isn't a closed form, so:
Attempt 2:
Using generating functions, there are $2n$ different balls and $n$ bins and we want each bin to hold exactly two balls.
So the generating functions would be: $(\frac {x^2} 2)^n$, so the closed form of that would be: $\frac 1 2 2^n \cdot n! = 2^{n-1}n!$ ?
In your Attempt 1 you should replace the $\sum$ by a $\prod$: One can select $n$ nominees from the list of $2n$ people in ${2n\choose n}$ ways. Order the chosen nominees alphabetically. The first nominee can choose his two voters in ${2n\choose 2}$ ways, then the second nominee can choose his two voters among the $2n-2$ left over voters in ${2n-2}$ ways, and so on. All in all we obtain $$N={2n\choose n}{2n\choose 2}{2n-2\choose 2}{2n-4\choose 2}\cdots{2\choose 2}={(2n)!\cdot(2n)!\over n!\>n!\>2^n}$$ possibilities. Here it was assumed that any voter is allowed to vote for himself.