How to divide 2n members of the club into disjoint teams of 2 members each, when teams not labelled? I am doing this and getting $\sum_{i=0}^{2n-2} \frac{(2n-i)!}{2!(2n-i-2)!}$ but the answer given is $\frac{(2n)!}{2^n n!}$ I even tried to get this by simplifying my result, but not getting the same.

Their answer: imagine lining up everyone in a row and then pairing up adjacent people. There are $(2n)!$ ways to line everyone up, but different line-ups can produce the same teams. To account for this overcounting, divide by $n!$ for the number of ways to arrange the $n$ teams in a row, and divide by $2$ for each pair to account for swapping the order of the members of each pair.
In your answer, you probably meant $\prod_{i=0}^{2n-2} \binom{2n-i}{2} = \prod_{i=0}^{2n-2}\frac{(2n-i)(2n-i-1)}{2} = \frac{(2n)!}{2^n}$, which comes from choosing the first pair, choosing the second pair, and so on. But here, you are accounting for the order that the teams are chosen. (Recall the problem states the teams are not labeled.) So, you should divide by $n!$. This agrees with their answer.
