Permutation of $n$ women and $m$ men, in a line, where the women dont get along with each other

Question

So the $n$ women can't sit next to each other. So in a straight line how many ways can they be seated?

I know this problem is partitioning distinct balls in $n+1$ partitions, out of which $n-1$ of the partitions should always be filled (the gap between the women) and multiply with $m! n!$

Assume $m \gt n$

Answer 1

Hint:

Arrange Men first and Then fill gaps between them with women

This method will work because No matter what there will be at least one man between them.

Don't forget to include spaces before first and after last man