11. A gardener randomly arranges three maple trees, four oak trees, and five birch trees in a row. Each different arrangement is equally likely.

Let $\mathrm{m}$ represent the probability that no two birch trees are adjacent (expressed as a simplified fraction). Find $\mathrm{m}+\mathrm{n}$.
There are 12! different arrangements of 12 trees. Let $\mathrm{k}$ be the number of arrangements where no two birch trees are adjacent. The required probability is $\frac{\mathrm{k}}{12!}$. Now, we need to find $\mathrm{k}$.

Using $t$ $\mathrm{N}$ to represent non-birch trees (maple or oak), we have (1)N (2)N (3)N (4)N (5)N (6)N (7)N (8)N, with 8 gaps "..." where 5 birch trees can be placed. The arrangement of the 7 non-birch trees has 7! ways, and for each arrangement, there are $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$ ways to arrange the birch trees. Therefore, $\mathrm{k}=(7!) 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$.
The required probability $\frac{\mathrm{k}}{12!}=(7!) \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{12!}$
$$
\begin{array}{l}
=\frac{7}{99} \text {, i.e., } \frac{\mathrm{m}}{\mathrm{n}}=\frac{7}{99} \\
\therefore \mathrm{m}+\mathrm{n}=106 .
\end{array}
$$