7・9 A gardener is to plant three maple trees, four oak trees, and five birch trees in a row. He randomly determines the order of these trees, with all different arrangements being equally probable. Let $\frac{m}{n}$ represent the probability that no two birch trees are adjacent (in simplest form), find $m+n$.


［Solution］There are 12! different arrangements for 12 trees.
There are 7 non-birch trees (3 fir trees and 4 oak trees). To ensure that no two birch trees are adjacent, the 5 birch trees can be placed in the gaps between the 7 non-birch trees. Since 7 non-birch trees have 8 gaps, the number of ways to arrange the 5 birch trees is
$$
p_{8}^{5}=8 \times 7 \times 6 \times 5 \times 4 \text { (ways). }
$$

The number of ways to arrange the 7 non-birch trees is $p_{7}^{7}=7!$ (ways).
Therefore, the number of arrangements where no two birch trees are adjacent is
$$
7!\cdot 8 \times 7 \times 6 \times 5 \times 4 \text { (ways). }
$$

Thus, the probability is
$$
p=\frac{7!\cdot 8 \times 7 \times 6 \times 5 \times 4}{12!}=\frac{7}{99} .
$$

So, $\frac{m}{n}=\frac{7}{99}$.
$$
m+n=99+7=106 .
$$