Difference between revisions of "2008 AIME I Problems/Problem 9"
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For (5,2,3),(3,5,2), we see that there are <math>2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7</math> ways to stack the crates. For (1,8,1), there are <math>2\dbinom{10}{2} = 90</math>. Also, there are <math>3^{10}</math> total ways to stack the crates to any height. | For (5,2,3),(3,5,2), we see that there are <math>2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7</math> ways to stack the crates. For (1,8,1), there are <math>2\dbinom{10}{2} = 90</math>. Also, there are <math>3^{10}</math> total ways to stack the crates to any height. | ||
Thus, our probability is <math>\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}</math>. Our answer is the numerator, <math>\boxed{190}</math>. | Thus, our probability is <math>\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}</math>. Our answer is the numerator, <math>\boxed{190}</math>. | ||
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+ | == See also == | ||
+ | {{AIME box|year=2008|n=I|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:50, 27 June 2014
Problem
Ten identical crates each of dimensions ft ft ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let be the probability that the stack of crates is exactly ft tall, where and are relatively prime positive integers. Find .
Solution
Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
Subtracting 3 times the third from the first gives , or . The last doesn't work, obviously. This gives the three solutions . In terms of choosing which goes where, the first two solutions are analogous.
For , we see that there are ways to stack the crates. For , there are . Also, there are total ways to stack the crates to any height.
Thus, our probability is . Our answer is the numerator, .
Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
$
Subtracting 3 times the second from the first gives , or (b,c) = (2,3),(5,2),(8,1),(11,0). The last doesn't work, obviously. This gives the three solutions (a,b,c) = (5,2,3),(3,5,2),(1,8,1). In terms of choosing which goes where, the first two solutions are analogous. For (5,2,3),(3,5,2), we see that there are ways to stack the crates. For (1,8,1), there are . Also, there are total ways to stack the crates to any height. Thus, our probability is . Our answer is the numerator, .
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.