I know that the partition of $2n$ items into $n$ pairs has something to do with double factorial, but I am not sure how many possibilities we exactly have.
We can choose such a partition into pairs by first choosing $n$ elements to be in one half and the remaining $n$ in the other half, which gives ${{2n} \choose n}$ possibilities. Then we can take a permutation of both sets and match them accordingly.
This gives $${{2n} \choose n} n! n!=\frac{(2n)! n! n!}{n! n!}=(2n)!$$ possibilities. But with this we overcount, since we do not want the pairs to be ordered. Each of the $n$ pairs is counted twice, wherefore we have to divide by 2 for every pair, hence $\frac{(2n)!}{2^n}$?
Well, another possiblity to do that is just to iteratively pick $2$ elements among the remaining ones, which gives ${{2n} \choose 2} {{2n-2} \choose 2} \dotsb {4 \choose 2}$
I am not sure, but I do not think that this is the same and also, I do not see the connection to double factorial. Would be very glad if you could help me with these 3 possiblities.
Thank you.