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:

So you can think of placing the $5$ birch trees and the other trees with the restrictions as described above. Then let's take out one tree between each pair of birch trees. So you would remove $4$ trees that aren't birch. What you are left with is a unique arrangement of $5$ birch trees and $3$ other trees that is unrestricted.

Some birch trees might become adjacent after you remove $4$ trees. Adding a tree between each pair of people gives a unique arrangement of $5$ nonadjacent birch trees. This guarantees that there are no adjacent trees. The number of unrestricted $7$ tree arrangments is ${8\choose5} = 56$.

There are ${12\choose5} = 792$ total ways to choose the trees. So the probability is $\frac{56}{792} = \frac{7}{99}$

This means our answer is $7 + 99 = \boxed{106}$