Ways to arrange $n\geq2$ people around a circular table, given two permanent seats.

Question

How many ways to arrange $n\geq2$ people around a circular table, given two specific people who cannot stand next to each other?

I've observed that when $n=2$ and $n=3$ there exists no way to arrange them so that the two specific people aren't standing next to each other. I think for when $n=4$ there are $3!$ ways to arrange them, but I've had trouble coming up with a formula that can describe all values of $n$.

Perhaps I can attack the problem by first arranging the other people in the circle, and then placing either one of the two specific people in between the gaps?

Answer 1

We can fix one of the 2 pearson. Then we have $n-3$ positions for the second pearson (since one is already ocupied and it can not be seated near to it), and then we can arrange the other on $(n-2)!$ ways, so the answer is $(n-3)\cdot (n-2)!$

Answer 2

Well there are $n$ seats. Assuming the seats are not numbered, so the first mandatory person can be placed anywhere.

Then, place the second mandatory person, there are $n-3$ choices (why?)

Place the rest of $n-2$ people into $n-2$ seats, so $(n-2)!$ ways

Total: $(n-3)(n-2)!$ ways