Just need to check if I understand this concept:

In how many different ways can we arrange 2n people into n pairs?

The equivalence classes for this problem would contain $\displaystyle 2^n$ ways to rearrange the pairs and n! ways to order the pairs, correct?

That would leave $\displaystyle \frac{(2n)!}{(2^{n})(n!)}$ as the answer.

I'm having a little trouble grasping the intricacies of partitioning and equivalence classes.