1984 AIME Problems/Problem 11

Revision as of 23:16, 5 April 2012 by Baijiangchen (talk | contribs) (Solution)


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 $m+n$.


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
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