34. (1984 American Invitational Mathematics Examination) A gardener is to plant three maple trees, four oaks, and five birch trees in a row. He randomly determines the order of these trees, with all different arrangements being equally likely. Express the probability that no two birch trees are adjacent as $\frac{m}{n}$ (in simplest form), and find $m+n$.


34.12 The different arrangements of 12 trees total 12 !.

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 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 ! $-8 \times 7 \times 6 \times 5 \times 4$ (ways).
Thus, the probability is
$$
P=\frac{7!\cdot 8 \times 7 \times 6 \times 5 \times 4}{12!}=\frac{7}{99} .
$$

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