2008 AIME I Problems/Problem 9
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:
$$ (Error compiling LaTeX. Unknown error_msg)\begin{align*}3a + 4b + 6c &= 41
a + b + c &= 10\end{align*}
b + 3c = 11
(b,c) = (2,3),(5,2),(8,1),(11,0)
(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$. In terms of choosing which goes where, the first two solutions are analogous.
For$ (Error compiling LaTeX. Unknown error_msg)(5,2,3),(3,5,2)
2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7
(1,8,1)
2\dbinom{10}{2} = 90
3^{10}$total ways to stack the crates to any height.
Thus, our probability is$ (Error compiling LaTeX. Unknown error_msg)\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{190}{3^{7}}
\boxed{190}$.
See also
| 2008 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
