Consider 10 people sitting around a circular table. In how many different ways can they change seats so that each person has a different neighbor to the right?
I'm not sure about my answer. It is an exercise from the book The Art and Craft of Problem Solving. My thought is to fix 1, while consider its left side, can not be 10, so it has 8 choices, and count them left to left, we get 8!. While if the final one is 2, then it is incorrect. Eliminate this situation, I think the answer is 8!-7!.
Solving this for smaller number of people, I get $a_1=0$, $a_2=0$, $a_3=1$ (132), $a_4=1$ (1432), and $a_5=8$ (13254, 13524, 13542, 14352, 14253, 15243, 15324, 15324, 15432).
Sticking those into OEIS, I got https://oeis.org/A000757, which gives us $a_{10}=120288$. It suggests that a recurrence relation is
$$a_n = (n-2) a_{n-1} + (n-1) a_{n-2} - (-1)^n$$ which the interested reader could verify.