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

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== Problem ==
 
== 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 <math>\frac m n</math> in lowest terms be the probability that no two birch trees are next to one another. Find <math>\displaystyle m+n</math>.
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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 <math>\frac m n</math> in lowest terms be the [[probability]] that no two birch trees are next to one another. Find <math>\displaystyle m+n</math>.
  
 
== Solution ==
 
== Solution ==
{{solution}}
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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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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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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>\displaystyle m+n = 106</math>.
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== See also ==
 
== See also ==
* [[1984 AIME Problems/Problem 10 | Previous problem]]
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{{AIME box|year=1984|num-b=10|num-a=12}}
* [[1984 AIME Problems/Problem 12 | Next problem]]
 
* [[1984 AIME Problems]]
 

Revision as of 22:53, 5 March 2007

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 $\displaystyle m+n$.

Solution

First off, notice that there are ${7\choose4} = 35$ ways of arranging just the maple and oak trees while ignoring the 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 ${8\choose5} = 56$ different ways to arrange this. Thus in total, there are $35 \cdot 56 = 1960$ such arrangements.

There are a total of $\frac{(3 + 4 + 5)!}{3! 4! 5!} = 27720$ ways of arranging the trees, so the requested probability is $\frac{1960}{27720} = \frac{7}{99}$, and $\displaystyle m+n = 106$.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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