5. Let $P$ be a convex $n\left(n \in \mathbf{N}_{+}, n>7\right)$-sided polygon. It is known that non-intersecting diagonals in $P$ divide $P$ into $n-2$ triangles. If one of these triangles has all three sides as diagonals of $P$, then it is called an "internal triangle". Try to find the number of ways to divide $P$ such that there are exactly two internal triangles (express the result in terms of $n$).
5. Label the vertices of $P$ in counterclockwise order as $A_{0}, A_{1}, \cdots, A_{n-1}$.

First, consider the number of partition methods in which two inner triangles are located as shown in Figure 3. Such a partition is said to start from point $A_{0}$.
In Figure 3, the numbers $m_{1}, m_{2}$ represent the number of sides of $P$ in region $M$, and $n_{1}, n_{2}, n_{3}, n_{4}$ represent the number of sides of $P$ in regions $N_{1}, N_{2}, N_{3}, N_{4}$, respectively. These regions are composed of "outer triangles" (i.e., triangles different from the inner triangles). The two inner triangles are $\triangle A_{0} A_{n_{1}} A_{n_{1}+n_{2}}$ and $\triangle A_{n_{1}+m_{2}+m_{1}} A_{n_{1}+n_{2}+m_{1}+n_{1}} A_{m_{1}+n_{2}+m_{1}+m_{3}+m_{4}}$.
We will prove that the partitions starting from point $A_{0}$ are in one-to-one correspondence with the seven-tuple $\left(m, n_{1}, n_{2}, n_{3}, n_{4}, w_{31}, w_{N}\right)$, where $m$ is a non-negative integer, $n_{1}, n_{2}, n_{3}, n_{4}$ are positive integers, and
\[ m+n_{1}+n_{2}+n_{3}+n_{4}=n, \]
\[ w_{n} \text{ is an } m \text{-bit binary number, and } w_{x} \text{ is an } (n-m-8) \text{-bit binary number.} \]

In fact, given a partition as shown in Figure 3, the number $m=m_{1}+m_{2}$ and $n_{1}, n_{2}, n_{3}, n_{4}$ satisfy equation (1) and $m \geqslant 0, n_{1}, n_{2}, n_{3}, n_{4} \geqslant 2$.

On the one hand, the partition method of region $N_{1}$ can uniquely determine an $(n_{1}-2)$-bit binary number as follows: Let the unique outer triangle in region $N_{1}$ with side $A_{0} A_{m_{1}}$ be $T_{1}$.

If $n_{1} \geqslant 3$, then the outer triangle $T_{1}$ has a unique adjacent outer triangle in region $N_{1}$, denoted as $T_{2}$; if $n_{1} \geqslant 4$, then the outer triangle $T_{2}$ has another adjacent outer triangle in region $N_{1}$, denoted as $T_{3}, \cdots \cdots$. Thus, we obtain a sequence of $n_{1}-1$ outer triangles in region $N_{1}$.

We assign values to this sequence as follows: If the third vertex of the outer triangle $T_{1}$ is $A_{1}$, then assign 0; if the third vertex of the outer triangle $T_{1}$ is $A_{n_{1}-1}$, then assign 1. In each case, the second vertex of the outer triangle $T_{2}$ has two possible choices, and when the third vertex has a smaller index, assign 0; when the third vertex has a larger index, assign 1. Continue this process to obtain a sequence of length $n_{1}-2$ consisting of 0s and 1s, where each term uniquely corresponds to the choice of the third vertex of the outer triangles $T_{1}, T_{2}, \cdots, T_{n_{1}-2}$. The third vertex of the outer triangle $T_{n_{1}-1}$ is uniquely determined by the choice of the outer triangle $T_{n_{1}-2}$. Thus, we obtain an $(n_{1}-2)$-bit binary number that uniquely corresponds to a partition of region $N_{1}$, i.e., there are $2^{n_{1}-2}$ such partition methods. Similarly, region $N_{i}$ corresponds to a binary number of length $(n_{i}-2)$, and the corresponding partition methods are $2^{n_{i}-2} (i=1,2,3,4)$.

Therefore, there exists an
\[ (n_{1}-2) + (n_{2}-2) + (n_{3}-2) + (n_{4}-2) = n-m-8 \]
bit binary number $w_{N}$, which corresponds one-to-one to the partition methods of regions $N_{1}, N_{2}, N_{3}, N_{4}$ (in fact, it corresponds to the partition methods of $P$ determined by $m_{1}, m_{2}, n_{1}, n_{2}, n_{3}, n_{4}$).

Next, we discuss the partition methods of region $M$.
Let the unique outer triangle in region $M$ with side $A_{0} A_{n_{1}+n_{2}}$ be $S_{1}$: if $m \leqslant 2$, then the outer triangle $S_{1}$ has a unique adjacent outer triangle in region $M$, denoted as $S_{2}$; if $m \geqslant 3$, then the outer triangle $S_{2}$ has another adjacent outer triangle in region $M$, denoted as $S_{3}, \cdots$. Thus, we obtain a sequence of $m$ outer triangles in region $M$.

We assign values to this sequence as follows: If the third vertex of the outer triangle $S_{1}$ is $A_{n_{1}+n_{2}+1}$, then assign 0; if the third vertex of the outer triangle $S_{1}$ is $A_{n-1}$, then assign 1. In each case, the third vertex of the outer triangle $S_{2}$ has two possible choices, and when the third vertex has a smaller index, assign 0; when the third vertex has a larger index, assign 1. Continue this process to obtain a sequence of length $m$ consisting of 0s and 1s, where each term uniquely corresponds to the choice of the third vertex of the outer triangles $S_{1}, S_{2}, \cdots, S_{n_{1}-2}$. Thus, we obtain an $m$-bit binary number that uniquely corresponds to a partition of region $M$, i.e., there are $2^{m}$ such partition methods. In particular, the number of 0s and 1s in such a binary number respectively correspond to $m_{1}$ and $m_{2}$, i.e., such a binary number can uniquely determine the values of $m_{1}$ and $m_{2}$. Therefore, given $n_{1}$ and $n_{2}$, the position of region $M$ in $P$ can be determined, and thus the positions of all regions in the figure can be determined.
And equation (D) satisfies
\[ m \geqslant 0, n_{1}, n_{2}, n_{3}, n_{4} \geqslant 2 \]

The number of integer solutions to the equation
\[ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = n-3 \]

is equivalent to the number of positive integer solutions to the equation, which is $\mathrm{C}_{n-4}^{4}$. Therefore, the number of seven-tuples $\left(m, n_{1}, n_{2}, n_{3}, n_{4}, w_{n}, w_{N}\right)$ (i.e., the number of partitions starting from point $A_{0}$) is
\[ 2^{m} \times 2^{n-m-8} \mathrm{C}_{n-4}^{4} = 2^{n-8} \mathrm{C}_{n-4}^{4}. \]

Since the partitions can start from any vertex of $P$, and each partition is counted twice when starting from two different vertices, the number of partitions that satisfy the condition is
\[ \frac{1}{2} \times 2^{n-8} n \mathrm{C}_{n-4}^{4} = 2^{n-9} n \mathrm{C}_{n-4}^{4}. \]