[Math] Number of ways to pair 2n elements from two different sets

Question

Say I have a group of 20 people, and I want to split them to pairs, I know that the
number of different ways to do it is $\frac{(2n)!}{2^n \cdot n!}$
But let's say that I have to pair a boy with a girl?
I got confused because unlike the first option the number of total elements (pairs) isn't $(2n)!$, and I failed to count it. I think that the size of each equivalence class is the same – $2^n \cdot n!$, I am just missing the number of total elements.
Any ideas?
Thanks!

Answer 1

If you have $n$ boys and $n$ girls, give them each a number and sort the pairs by wlog the boy's number. Then there are $n!$ possible orderings for the girls, so $n!$ ways of forming the pairs.