Difference between revisions of "2009 AIME I Problems/Problem 10"
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== 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 | ||
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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: | 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: | ||
Revision as of 01:59, 15 February 2014
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 
to 
in clockwise order. Committee rules state that a Martian must occupy chair and an Earthling must occupy chair , 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 
. Find 
.
Solution
since the 5 members of each plant committee are distinct we get that the number of arrangement of sittings is in the form 
because for each M,V,E sequence we have 
arrangements within the Ms, Vs, and Es. which results in a number above
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. One block of five people- There is only one way to arrange this so 
.
2. Five blocks of one person - There is also only one way to arrange this so we get .
3. Two blocks - There are two cases: 
and 
. Each of these can be arranged two ways so we get 
.
4. Three blocks - There are also two cases: 
and 
.Each of these can be arranged three ways giving us 
.
5. Four blocks - There is only one case: 
. This can be arranged four ways giving us 
.
Combining all these cases, we get 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
