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.

Look at the problem of arranging letters in the word MAMA. It is $\binom{4}{2}$,  because it enumerates all ways to get 2 disjoint subsets froma set of 4 numbers and compare to the case when Ms and As are of different colors: you need to mulfiply it by 2!. The first case is your situation.
