7.9 A gardener is going to plant three maple trees, four oak trees, and five birch trees in a row. He randomly determines the order of these trees, and all different arrangements are equally probable. Use $\frac{m}{n}$ to represent the probability that no two birch trees are adjacent (simplified to the lowest terms), and find $m+n$.


[Solution] There are 12! different arrangements of 12 trees.
There are 7 non-birch trees (3 maple 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 there are 8 gaps among the 7 non-birch trees, 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
$$7!\cdot 8 \times 7 \times 6 \times 5 \times 4 \text { (ways) } \text {. }$$

Thus, the probability is
$$\begin{array}{l}
p=\frac{7!\cdot 8 \times 7 \times 6 \times 5 \times 4}{12!}=\frac{7}{99} . \\
\frac{m}{n}=\frac{7}{99} . \\
m+n=99+7=106 .
\end{array}$$
$$\text { That is, } \frac{m}{n}=\frac{7}{99} \text {. }$$