2009 AIME I Problems/Problem 10

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 $1+1+64+64+216= \boxed{346}$

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

Page

Toolbox

Search

2009 AIME I Problems/Problem 10

Problem

Solution

See also