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

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 $7 + 99 = \boxed{106}$.