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Difference between revisions of "2008 AMC 10B Problems/Problem 21"

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==Solution 2==
 
==Solution 2==
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Label the seats ABCDEFGHIJ, where A is the top seat. The first man has <math>10</math> possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are <math>4! = 24</math> ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be <math>10\cdot 2\cdot 24\cdot x = 480x</math> possible seating arrangements, where x is a nonnegative integer. There is only one answer that is a multiple of <math>480</math>. So, our answer is <math>\boxed{(\text{C}) 480}</math>. -Milk_123
+
Label the seats ABCDEFGHIJ, where A is the top seat. The first man has <math>10</math> possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are <math>4! = 24</math> ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be <math>10\cdot 2\cdot 24\cdot x = 480x</math> possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of <math>480</math>. So, our answer is <math>\boxed{(\text{C}) 480}</math>. -Milk_123
  
 
==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 21:35, 6 November 2021

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 $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{(\text{C}) 480}$.

Solution 2

Label the seats ABCDEFGHIJ, where A is the top seat. The first man has $10$ possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are $4! = 24$ ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be $10\cdot 2\cdot 24\cdot x = 480x$ possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of $480$. So, our answer is $\boxed{(\text{C}) 480}$. -Milk_123

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

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