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2009 AIME I Problems/Problem 10

Revision as of 16:23, 17 March 2018 by Mingxu (talk | contribs) (→‎Solution)

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 the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es.

Pretend the table only seats $3$ "people", with $1$ "person" from each planet. Counting clockwise, only the arrangement M, V, E satisfies the given constraints. Therefore, in the actual problem, the members must sit in cycles of M, V, E, but not necessarily with one M, one V, and one E in each cycle(for example, MMVVVE, MVVVEEE, MMMVVVEE all count as cycles). These cycles of MVE must start at seat $1$, since an M is at seat $1$. We simply count the number of arrangements through casework.

1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE

2. Two cycles - There are 3 Ms, Vs and Es left to distribute among the existing MVEMVE. Using sticks and stones(stars and bars), we get $\binom{4}{1}=4$ ways for the members of each planet. Therefore, there are $4^3=64$ ways in total.

3. Three cycles - 2 Ms, Vs, Es left, so $\binom{4}{2}=6$, making there $6^3=216$ ways total.

4. Four cycles - 1 M, V, E left, each M can go to any of the four MVE cycles and likewise for V and E, $4^3=64$ ways total

5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just $1$ way.

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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