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 $m+n=106.$

1984 AIME Problem 11

Answer: 106

Difficulty rating: 2160

Small Hint:

Big Hint:

Solution:

Problem 11 in Other Years