Difference between revisions of "2008 AIME I Problems/Problem 9"
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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>. | ||
| − | + | 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}. |
Revision as of 19:12, 1 April 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: \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}.
