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

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?

• @saulspatz Thanks for the info. Updated the body with what I've come up with so far. – ivyleaf57 Feb 6 at 17:51
• Just think of the number of ways you can seat the two next to each other for $N$ people. Subtract that from the total number of ways to arrange $N$ people. – Bridgeburners Feb 6 at 18:01
• Does rotated things count? And reflected? – Math Lover Feb 6 at 18:15

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 $$(n-3)\cdot (n-2)!$$
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