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

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== Problem ==
 
== 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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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 \cdot (5!)^3</math>.  Find <math>N</math>.
  
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== Solution ==
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== Solution 1 ==
 
Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form <math>N*(5!)^3</math> because for each <math>M, V, E</math> sequence we have <math>5!</math> arrangements within the Ms, Vs, and Es.
 
Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form <math>N*(5!)^3</math> because for each <math>M, V, E</math> sequence we have <math>5!</math> arrangements within the Ms, Vs, and Es.
  
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1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE
 
1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE
  
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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 <math>\binom{4}{1}=4</math> ways for the members of each planet. Therefore, there are <math>4^3=64</math> ways in total.
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2. Two cycles - There are 3 Ms, Vs and Es left to distribute among the existing MVEMVE. Using stars and bars, we get <math>\binom{4}{1}=4</math> ways for the members of each planet. Therefore, there are <math>4^3=64</math> ways in total.
  
 
3. Three cycles - 2 Ms, Vs, Es left, so <math>\binom{4}{2}=6</math>, making there <math>6^3=216</math> ways total.
 
3. Three cycles - 2 Ms, Vs, Es left, so <math>\binom{4}{2}=6</math>, making there <math>6^3=216</math> ways total.
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Combining all these cases, we get <math>1+1+64+64+216= \boxed{346}</math>
 
Combining all these cases, we get <math>1+1+64+64+216= \boxed{346}</math>
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== Solution 2 ==
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The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for <math>1, 2, 3, 4,</math> or <math>5</math> MVE's. If there are <math>k</math> MVE's, then by stars and bars, there are <math>{4 \choose k-1}</math> choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have <math>N = 1^3+4^3+6^3+4^3+1^3 = \boxed{346}</math> - aops5234
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== Note ==
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You are probably confused by the solutions. I was too when I first read them. Here is the intuition:
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Consider the the arrangement of the <math>M</math>'s <math>V</math>'s and <math>E</math>'s: <math>MMMMM</math> | <math>VVVVV</math> | <math>EEEEE</math>. Now, we can place a bar in the first Martian's section in <math>4</math> spots, and same with the Venusians and Earthlings.
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Now for instance, <math>MM|M|MM</math>    ||    <math>V|V|VVV</math>    ||    <math>EEE|E|E</math> would represent the table ordering MMVEEE MVE MMVVVE, which is a valid ordering. Hence we proceed with the solution and we are done.
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~th1nq3r
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== Video Solution  ==
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https://youtu.be/fNtw1PKE67w?si=6N8_pWHWnURgxkGv
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 +
~MathProblemSolvingSkills
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== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2009|n=I|num-b=9|num-a=11}}
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[[Category: Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:49, 27 September 2023

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 $N \cdot (5!)^3$. Find $N$.

Solution 1

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

Solution 2

The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for $1, 2, 3, 4,$ or $5$ MVE's. If there are $k$ MVE's, then by stars and bars, there are ${4 \choose k-1}$ choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have $N = 1^3+4^3+6^3+4^3+1^3 = \boxed{346}$ - aops5234

Note

You are probably confused by the solutions. I was too when I first read them. Here is the intuition: Consider the the arrangement of the $M$'s $V$'s and $E$'s: $MMMMM$ | $VVVVV$ | $EEEEE$. Now, we can place a bar in the first Martian's section in $4$ spots, and same with the Venusians and Earthlings. Now for instance, $MM|M|MM$ || $V|V|VVV$ || $EEE|E|E$ would represent the table ordering MMVEEE MVE MMVVVE, which is a valid ordering. Hence we proceed with the solution and we are done.

~th1nq3r


Video Solution

https://youtu.be/fNtw1PKE67w?si=6N8_pWHWnURgxkGv

~MathProblemSolvingSkills


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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