2008 AMC 10B Problem 21

Because the chairs are numbered, rotations count as different arrangements.

Men and women alternate, so the men occupy either all the odd chairs or all the even chairs: $2$ choices. The five men can then be seated in $5! = 120$ ways.

Now fix the men in chairs $1, 3, 5, 7, 9$ (the even case is identical). The wife of the man in chair $k$ may not sit in chairs $k-1$, $k+1$ or $k+5$ (across), all of which are even, so she has $5 - 3 = 2$ legal chairs:
$$W_1: \{4, 8\},\quad W_3: \{6, 10\},\quad W_5: \{2, 8\},\quad W_7: \{4, 10\},\quad W_9: \{2, 6\}.$$
If $W_1$ takes chair $4$, then $W_7$ must take $10$, so $W_3$ takes $6$, so $W_9$ takes $2$, so $W_5$ takes $8$. If $W_1$ takes chair $8$, then $W_5$ takes $2$, $W_9$ takes $6$, $W_3$ takes $10$, $W_7$ takes $4$. Every step is forced, so there are exactly $2$ ways to seat the wives.

Total: $2 \cdot 120 \cdot 2 = 480$.

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

Seat the less constrained group first, then treat the other group as an assignment problem. Because each wife has exactly two options and each chair is wanted by exactly two wives, the options form a single closed chain, and a chain like that admits exactly two perfect assignments. Which man sits where is irrelevant by symmetry: the forbidden pattern depends only on chair numbers.

• Dividing by 10 for rotations even though the chairs are numbered, or assuming each wife independently has 2 choices and multiplying by 2^5.
• Forgetting the "across" restriction, which would give each wife $3$ options and a larger count.