Difference between revisions of "1984 AIME Problems/Problem 11"

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== Solution ==
 
== Solution ==
First off, notice that there are <math>{7\choose4} = 35</math> ways of arranging just the maple and oak trees while ignoring the birch trees.
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First notice that there is no difference between the maple trees and the oak trees; we have only two types, birth trees and "non-birch" trees.
  
The five birch trees must now be placed amongst the seven previous trees. We can think of these trees as 7 dividers of 8 slots that the birch trees can go in, making <math>{8\choose5} = 56</math> different ways to arrange this. Thus in total, there are <math>35 \cdot 56 = 1960</math> such arrangements.
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The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 7 dividers of 8 slots that the birch trees can go in, making <math>{8\choose5} = 56</math> different ways to arrange this.
  
There are a total of <math>\frac{(3 + 4 + 5)!}{3! 4! 5!} = 27720</math> ways of arranging the trees, so the requested probability is <math>\frac{1960}{27720} = \frac{7}{99}</math>, and <math>m+n = 106</math>.
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There are <math>{12 \choose 5} = 792</math> total ways to arrange the twelve trees, so the probability is <math>\frac{56}{792} = \frac{7}{99}</math>.
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The answer is <math>7 + 99 = \boxed{106}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:16, 5 April 2012

Problem

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$.

Solution

First notice that there is no difference between the maple trees and the oak trees; we have only two types, birth trees and "non-birch" trees.

The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 7 dividers of 8 slots that the birch trees can go in, making ${8\choose5} = 56$ different ways to arrange this.

There are ${12 \choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\frac{56}{792} = \frac{7}{99}$.

The answer is $7 + 99 = \boxed{106}$.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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