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Difference between revisions of "2009 AIME I Problems/Problem 10"
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== Solution ==
 
== 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:
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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. These blocks must be of the form MVE with a certain number of M's,V's,and E's,we consider a few different cases:
  
 
1. One block of five people- There is only one way to arrange this so <math>{1^3}=1</math>.
 
1. One block of five people- There is only one way to arrange this so <math>{1^3}=1</math>.

Revision as of 08:14, 2 June 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. These blocks must be of the form MVE with a certain number of M's,V's,and E's,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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