1984 AIME Problem 11

Question

A gardener plants three maple trees, four oak trees, 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.

Answer

Answer: 106

Concepts:arrangements with restrictions combinations basic probability

Difficulty rating: 2160

Small hint:

First choose the five positions occupied by birch trees

Big hint:

Place the seven non-birch trees first and use the eight gaps around them

Solution:

The five birch positions form a uniformly chosen

5

-element subset of the

12

positions, so there are

\binom{12}{5}

possibilities. After the seven non-birch trees are placed, there are eight gaps, including the two end gaps. Choosing five distinct gaps gives

\binom85

arrangements with no adjacent birches. Therefore

\frac{m}{n}=\frac{\binom85}{\binom{12}{5}} =\frac{56}{792}=\frac7{99},

and

m+n=106.