Two round table combinatorics problems. I am struggling to answer the following two problems: How many ways can 4 men and 8 women be seated at a round table if there are to be two women and between each man? How many ways can 15 people be seated at a round table if B refuses to sit next to A?

Let the men be A, B, C, D. In problems about circular permutations, two permutations that differ by a rotation are usually considered to be the same. Equivalently, we can assume that there is a special chair, and that A is sitting on it.

Then the set of _positions_ occupied by the men is determined, and the remaining men can be arranged in these in $3!$ ways. For each such way, the women can be arranged in $8!$ ways.

For the second problem, you will find the following approach useful. Call a placement **bad** if A and B are next to each other. Count the number of bad placements, and subtract this from the total number of possible placements.

Another way of solving the problem is to assume as before that A sits in the special chair. How many ways are there to seat B? And now how many ways are there to seat the rest?

**Remark:** It is sometimes a good idea to solve a problem in two different ways. That can provide a (partial) check of correctness.