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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 ==
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since the 5 members of each plant committee are distinct we get that the number of arrangement of sittings is in the form <math>N*(5!)^3</math> because for each M,V,E sequence we have <math>5!</math> arrangements within the Ms, Vs, and Es. which results in a number above
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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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Pretend the table only seats <math>3</math> "people", with <math>1</math> "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 <math>1</math>, since an M is at seat <math>1</math>. We simply count the number of arrangements through casework.
  
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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:
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1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE
  
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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. 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.
  
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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. 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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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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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, <math>4^3=64</math> ways total
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5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just <math>1</math> way.
  
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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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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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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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~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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Combining all these cases, we get <math>1+1+64+64+216= \boxed{346}</math>
 
  
 
== 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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