2008 AMC 10B Problem 21

Attempt Problem 21 of the 2008 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

Ten chairs are evenly spaced around a round table and numbered clockwise from 11 through 10.10. Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or directly across from his or her spouse. How many seating arrangements are possible?
240240
360360
480480
540540
720720
Answer: C
Concepts:circular arrangementsarrangements with restrictionscasework
Small hint:
Seat the women first; count their arrangements in the alternating seats
Big hint:
For a fixed seating of the women, count how many ways the men can avoid their spouses
Solution:
Seat the women first. The first woman may take any of the 1010 chairs, and since seats alternate, the remaining women fill their four seats in 4!4! ways, giving 10⋅4!=24010\cdot 4!=240 arrangements.
Fix a woman in chair 1.1. Her spouse must sit in chair 44 or chair 8;8; each choice then forces the placement of every other man consistently. So each seating of the women yields exactly 22 valid seatings of the men.
The total is 2⋅240=480.2\cdot 240=480.
Thus, the correct answer is C.
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