Difference between revisions of "2008 AMC 10B Problems/Problem 21"
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==Solution== | ==Solution== | ||
| − | For the first man, there are <math>10</math> possible seats. For each subsequent man, there are <math>4</math>, <math>3</math>, <math>2</math>, and <math>1</math> possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is <math>10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{(\text{C})480}</math>. | + | For the first man, there are <math>10</math> possible seats. For each subsequent man, there are <math>4</math>, <math>3</math>, <math>2</math>, and <math>1</math> possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is <math>10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{(\text{C}) 480}</math>. |
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:15, 31 January 2016
Problem
Ten chairs are evenly spaced around a round table and numbered clockwise from 
through 
. 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?
Solution
For the first man, there are possible seats. For each subsequent man, there are 
, 
, 
, and possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
