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Difference between revisions of "2008 AIME I Problems/Problem 9"
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a + b + c &= 10\end{align*}</cmath>
 
a + b + c &= 10\end{align*}</cmath>
  
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Subtracting 3 times the third from the first gives <math>b + 3c = 11</math>, or <math>(b,c) = (2,3),(5,2),(8,1),(11,0)</math>. The last doesn't work, obviously. This gives the three solutions <math>(a,b,c) = (5,2,3),(3,5,2),(1,8,1)</math>. In terms of choosing which goes where, the first two solutions are analogous.
+
Subtracting 3 times the second from the first gives <math>b + 3c = 11</math>, or <math>(b,c) = (2,3),(5,2),(8,1),(11,0)</math>. The last doesn't work, obviously. This gives the three solutions <math>(a,b,c) = (5,2,3),(3,5,2),(1,8,1)</math>. In terms of choosing which goes where, the first two solutions are analogous.
  
 
For <math>(5,2,3),(3,5,2)</math>, we see that there are <math>2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7</math> ways to stack the crates. For <math>(1,8,1)</math>, 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 <math>(5,2,3),(3,5,2)</math>, we see that there are <math>2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7</math> ways to stack the crates. For <math>(1,8,1)</math>, 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.

Revision as of 13:57, 5 June 2019

Problem

Ten identical crates each of dimensions $3$ ft $\times$ $4$ ft $\times$ $6$ 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 $\frac {m}{n}$ be the probability that the stack of crates is exactly $41$ ft tall, where $m$ and $n$ are relatively prime positive integers. Find $m$.

Solution

Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:

\begin{align*}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{align*}

Subtracting 3 times the second from the first gives $b + 3c = 11$, 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 $2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $(1,8,1)$, there are $2\dbinom{10}{2} = 90$. Also, there are $3^{10}$ total ways to stack the crates to any height.

Thus, our probability is $\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}$. Our answer is the numerator, $\boxed{190}$.

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

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