2008 AMC 10B Problems/Problem 21

Problem

Ten chairs are evenly spaced around a round table and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?

$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$

Solution

For the first man, there are $10$ possible seats. For each subsequent man, there are $4$ , $3$ , $2$ , or $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{(\text{C}) 480}$ .

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 10 Problems and Solutions

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2008 AMC 10B Problems/Problem 21

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