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Difference between revisions of "2009 AIME I Problems/Problem 10"

(New page: 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. Block of 5 - There is ...)
 
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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 ==
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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:
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1. Block of 5 - There is only one way to arrange this so 1^3=1
 
  
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2. 5 Blocks of 1 - There is also only one way to arrange this so we get 1^3=1
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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>.
  
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3. 2 Blocks - There are two cases: 4+1 and 3+2. Each of these can be arranged two ways so we get (2+2)^3=64
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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>.
  
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4. 3 Blocks - There are also two cases: 3+1+1 and 2+2+1.Each of these can be arranged 3 ways giving us (3+3)^3=216
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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>.
  
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5. 4 Blocks - There is only one case: 2+1+1+1. This can be arranged 4 ways giving us 4^3=64
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Combining all these cases, we get <math>1+1+64+64+216= \boxed{346}</math>
  
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Combining all these cases, we get 1+1+64+64+216=346
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== See also ==
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{{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 $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Earthling must occupy chair $15$, 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 $N(5!)^3$. Find $N$.

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 ${1^3}=1$.

2. Five blocks of one person - There is also only one way to arrange this so we get ${1^3}=1$.

3. Two blocks - There are two cases: $4+1$ and $3+2$. Each of these can be arranged two ways so we get ${(2+2)^3}=64$.

4. Three blocks - There are also two cases: $3+1+1$ and $2+2+1$.Each of these can be arranged three ways giving us ${(3+3)^3}=216$.

5. Four blocks - There is only one case: $2+1+1+1$. This can be arranged four ways giving us ${4^3}=64$.

Combining all these cases, we get $1+1+64+64+216= \boxed{346}$

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
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