2008 AMC 10B Problem 21

Below is the professionally curated solution for Problem 21 of the 2008 AMC 10B, 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.

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Concepts:circular arrangementsarrangements with restrictionscasework

Difficulty rating: 1870

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

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 104!=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 2240=480.2\cdot 240=480.

Thus, the correct answer is C.

Problem 21 in Other Years