# Partitioning elements into subsets

The main question I'm trying to solve is how many 1-factors are in $K_{2n}$ where $K_n$ is the complete graph on $n$ vertices.

Would it be correct to think of it as how many ways can you partition 2n elements into n subsets of size 2? Because that doesn't really make sense to me (it was an idea given to me by someone else).

For example,

when $n=2$ then the set of 2n = 4 elements, {$x_1,x_2,x_3,x_4$}, will have 6 subsets of size 2. But how would I determine how to partition the set into 2 subsets of size 2?

Overall, is this the right thought process? If not, how should I start solving this problem?

Now, lets think about how to count these. Suppose we choose our first subset of size two - there are $\binom{2n}{2}$ ways of doing this. Now, we have $2n-2$ elements left to partition. There are $\binom{2n-2}{2}$ ways of choosing the next subset of size 2 and so on. However, we do not distinguish which order these sets are in, so we should divide by $n!$. In total, there should $\frac{1}{n!}\prod_{j=1}^n\binom{2j}{2}$ ways to partition this set. Now, $\binom{2j}{2}=\dfrac{2j(2j-1)}{2}=j(2j-1)$. We can write $$\frac{1}{n!}\prod_{j=1}^n\binom{2j}{j}=\frac{1}{n!}\prod_{j=1}^nj(2j-1)=\frac{1}{n!}\prod_{j=1}^nj\prod_{j=1}^n(2j-1).$$ Now, the middle term is clearly $n!$. For the third term, we can recognize that with a little manipulation, this is $\frac{(2n)!}{2^nn!}$ since $2^nn!$ will be the product of all the even terms in $(2n)!$, leaving only the odd terms. Thus, since the first and second terms cancel, the total number of ways to 1-factors is $\frac{(2n)!}{2^nn!}$.