Difference between revisions of "2009 AIME I Problems/Problem 10"
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You are probably confused by the solutions. I was too when I first read them. Here is the intuition: | 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 <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. | 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. | ||
− | Now for instance, <math>MM|M|MM</math> | + | 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. |
~th1nq3r | ~th1nq3r |
Revision as of 13:46, 29 November 2021
Contents
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 1
Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form because for each
sequence we have
arrangements within the Ms, Vs, and Es.
Pretend the table only seats "people", with
"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
, since an M is at seat
. 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 ways for the members of each planet. Therefore, there are
ways in total.
3. Three cycles - 2 Ms, Vs, Es left, so , making there
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, ways total
5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just way.
Combining all these cases, we get
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 or
MVE's. If there are
MVE's, then by stars and bars, there are
choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have
- 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 's
's and
's:
|
|
. Now, we can place a bar in the first Martian's section in
spots, and same with the Venusians and Earthlings.
Now for instance,
||
||
would represent the table ordering MMVEEE MVE MMVVVE, which is a valid ordering. Hence we proceed with the solution and we are done.
~th1nq3r
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.