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>, | + | 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>, or <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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| + | ==Solution 2== | ||
| + | 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>. | ||
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| + | ~ 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}} | ||
Latest revision as of 21:37, 6 November 2021
Contents
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 
, 
, 
, or possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is 
.
Solution 2
Label the seats ABCDEFGHIJ, where A is the top seat. The first man has 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 
ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be 
possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of 
. So, our answer is 
.
~ 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
