How many ways can this be done? Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
I can see a way to solve this question by just listing but I was wondering if there is a more clever way. If not, what would be the best way to list here?
 A: To simplify a little you can consider the configuration where all the persons, after changing their places, go to the diametral opposite chair. So the rule becomes that each person can either stay where it was, or can move to an adjacent chair.
Let $1, 2, 3, 4, 5, 6$ be the places (modulo $6$). If $n$ goes into $n+1$, then $n+1$ can either go to $n$ or to $n+2$. But in the second case, $n+2$ is forced to go to $n+3...$ and so on. So the orbits of the permutation can have order $1$ (when one person stays where he is), $2$ (when two adjacent persons switch their places) or 6 (when all the persons rotate).
The $2$ rotations (clockwise and counter-clockwise) are the only permutations with order 6 orbits.
All the other permutation are identified by choosing the set of pairs of adjacent person which will swap (transpositions). These can be listed by the number of fixed points.
$6$ fixed points: everyone stays still, there is only one way to do this
$4$ fixed points: you can choose the edge of the only transposition in 6 different ways.
$2$ fixed points: they must have an odd distance: $1$ or $3$. Six configurations have two adjacent fixed points. Three configurations have two opposite fixed points.
$0$ fixed points: there are two configurations.
In total I have counted 20 possibilities: $(123456), (654321), (), (12), (23), (34), (45), (56), (61), (12)(34), (23)(45), (34)(56), (45)(61), (56)(12), (61)(23), (12)(45), (23)(56), (34)(61), (12)(34)(56), (23)(45)(61)$.
