2008 AMC 10B Problem 21

The chairs are numbered, so turning the whole table gives a different arrangement. Seat the men first, then see where the wives are allowed to go.

Seating the men.

1. Men and women alternate, so the men sit either in all the odd chairs or in all the even chairs: $2$ choices.
2. The $5$ men can be placed in their $5$ chairs in $5 \times 4 \times 3 \times 2 \times 1 = 120$ ways.

Seating the wives. Suppose the men are in chairs $1, 3, 5, 7, 9$ (the even case works the same way). Call the wife of the man in chair $k$ "wife $k$".

1. Wife $k$ cannot sit in the two chairs next to chair $k$, or in the chair straight across, which is $5$ chairs away. Of the five even chairs, that rules out three, leaving two:

1. Try wife $1$ in chair $4$. Chair $4$ is taken, so wife $7$ must take $10$. Then wife $3$ must take $6$, wife $9$ must take $2$, and wife $5$ must take $8$. Every step is forced: one seating.
2. Try wife $1$ in chair $8$. Then wife $5$ takes $2$, wife $9$ takes $6$, wife $3$ takes $10$, and wife $7$ takes $4$. Again forced: one seating.
3. So the wives can be seated in exactly $2$ ways, however the men sat.

Total. $2 \times 120 \times 2 = 480$.

The answer is $\boxed{\textbf{(C)}\ 480}$.

Seat one group first, then list where each person in the other group may go. Here every wife has two chairs and every chair is wanted by two wives, so the choices link together in one loop. Once the first wife picks, everyone else is forced, which gives only two ways, not $2 \times 2 \times 2 \times 2 \times 2$. The men's order does not matter, because the allowed chairs depend only on chair numbers.