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

m
(Solution)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
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
+
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.
  
 +
Pretend the table only sits 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 a M is at seat 1. We simply count the number of arrangements through casework.
  
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. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE
  
1. One block of five people- There is only one way to arrange this so <math>{1^3}=1</math>.
+
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 C(4,1)=4 ways for the members of each planet. Therefore, there are <math>4^3=64</math> ways in total.
  
2. 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>.
+
3. Three cycles - 2 Ms, Vs, Es left, so C(4,2)=6 and 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, 64 ways total
  
3. 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>.
+
5. Five cycles - MVEMVEMVEMVEMVE=1 way
 
 
4. 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>.
 
 
 
5. Five blocks of one person - There is also only one way to arrange this so we get <math>{1^3}=1</math>.
 
  
 
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>

Revision as of 21:12, 16 February 2015

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 plant 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 sits 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 a 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 C(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 C(4,2)=6 and 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, 64 ways total

5. Five cycles - MVEMVEMVEMVEMVE=1 way

Combining all these cases, we get $1+1+64+64+216= \boxed{346}$

See also

2009 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png