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.

Your analysis leads to the $\frac{2n!}{2^n}$ part of the answer, but overlooks the overcount due to the order of the groups.
As a direct method, consider simply starting with all orderings of the $2n$ members ($2n!$ options) and making groups by first $2$, next $2$ etc. This overcounts in two ways 


*

*the order of the members in the pairing

*the order of the pairs


Dividing out these overcounts gives the result $\dfrac{2n!}{2^nn!}$
