Difference between revisions of "2008 AMC 10B Problems/Problem 21"
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==Problem== | ==Problem== | ||
| − | + | Ten chairs are evenly spaced around a round table. 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? | |
| + | |||
| + | A) 240 B) 360 C) 480 D) 540 E) 720 | ||
==Solution== | ==Solution== | ||
| − | + | For the first man, there are 10 possible seats. For each subsequent man, there are 4, 3, 2, and 1 possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is 10*4*3*2*1*2 = 480 (C) | |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2008|ab=B|num-b=20|num-a=22}} | ||
Revision as of 12:58, 20 December 2008
Problem
Ten chairs are evenly spaced around a round table. 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?
A) 240 B) 360 C) 480 D) 540 E) 720
Solution
For the first man, there are 10 possible seats. For each subsequent man, there are 4, 3, 2, and 1 possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is 10*4*3*2*1*2 = 480 (C)
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
