How many $n$ women choose $n$ dance partners if everyone has to change partners for the second dance

Question

I am trying assignment problems of Combinatorics of an Institute in which I am not a student as our professor doesn't give any assignment.

I am struck on this problem and couldn't think about.

Problem is -> Suppose at a party there are $n$ men and $n$ women. In how many ways can the n women chose male partners for the first dance? How many ways are there for the second dance if everyone has to change partners?

For first part it is clear that solution would be $n!$. But I am not able to think how to approach problem for 2nd part as I must ignore a man for every women which she earlier chose.

Can anybody please explain how to solve it?

Answer 1

I suggest to look on the internet for "Derangements".