Difference between revisions of "2009 AIME I Problems/Problem 10"
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| + | == Problem == | ||
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| + | The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from <math>1</math> to <math>15</math> in clockwise order. Committee rules state that a Martian must occupy chair <math>1</math> and an Earthling must occupy chair <math>15</math>, Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is <math>N(5!)^3</math>. Find <math>N</math>. | ||
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| + | == Solution == | ||
| + | |||
Since after each planet, only members of another planet can follow, we simply count the lengths of the blocks adding up to ten. We consider a few different cases: | Since after each planet, only members of another planet can follow, we simply count the lengths of the blocks adding up to ten. We consider a few different cases: | ||
| − | |||
| − | 2. | + | 1. One block of five people- There is only one way to arrange this so <math>{1^3}=1</math>. |
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| + | 2. Five blocks of one person - There is also only one way to arrange this so we get <math>{1^3}=1</math>. | ||
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| + | 3. Two blocks - There are two cases: <math>4+1</math> and <math>3+2</math>. Each of these can be arranged two ways so we get <math>{(2+2)^3}=64</math>. | ||
| − | + | 4. Three blocks - There are also two cases: <math>3+1+1</math> and <math>2+2+1</math>.Each of these can be arranged three ways giving us <math>{(3+3)^3}=216</math>. | |
| − | + | 5. Four blocks - There is only one case: <math>2+1+1+1</math>. This can be arranged four ways giving us <math>{4^3}=64</math>. | |
| − | + | Combining all these cases, we get <math>1+1+64+64+216= \boxed{346}</math> | |
| − | + | == See also == | |
| + | {{AIME box|year=2009|n=I|num-b=9|num-a=11}} | ||
Revision as of 14:00, 20 March 2009
Problem
The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from 
to 
in clockwise order. Committee rules state that a Martian must occupy chair and an Earthling must occupy chair , Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is 
. Find 
.
Solution
Since after each planet, only members of another planet can follow, we simply count the lengths of the blocks adding up to ten. We consider a few different cases:
1. One block of five people- There is only one way to arrange this so 
.
2. Five blocks of one person - There is also only one way to arrange this so we get .
3. Two blocks - There are two cases: 
and 
. Each of these can be arranged two ways so we get 
.
4. Three blocks - There are also two cases: 
and 
.Each of these can be arranged three ways giving us 
.
5. Four blocks - There is only one case: 
. This can be arranged four ways giving us 
.
Combining all these cases, we get 
See also
| 2009 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
