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>.
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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 11:15, 31 January 2016

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

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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

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