United States — Team Selection Test

The answer is
$$n2^{n-9} \binom{n-4}{4}$$
Denote the vertices of $\mathcal{P}$ counter-clockwise by $A_0, A_1, \dots, A_{n-1}$. We will count first the number of triangulations of $\mathcal{P}$ with two interior triangles positioned as in the following figure. We say that such a triangulation starts at $A_0$.

The numbers $m_1, m_2, n_1, n_2, n_3, n_4$ in the figure denote the number of sides of $P$ determining the regions $N_1, N_2, N_3, N_4$ and $M$ that consist of exterior triangles (triangles that are not interior). The two interior triangles are
$$A_0 A_{n_1} A_{n_1+n_2} \text{ and } A_{n_1+n_2+m_1} A_{n_1+n_2+m_1+n_3} A_{n_1+n_2+m_1+n_3+n_4},$$
respectively.

We will show that triangulations starting at $A_0$ are in bijective correspondence to 7-tuples
$$(m, n_1, n_2, n_3, n_4, w_M, w_N),$$
where $m \ge 0$, $n_1, n_2, n_3, n_4 \ge 2$ are integers,
$$m + n_1 + n_2 + n_3 + n_4 = n, \qquad (\dagger)$$
$w_M$ is a binary sequence (sequence of 0's and 1's) of length $m$ and $w_N$ is a binary sequence of length $n - m - 8$.

Indeed, given a triangulation as in the figure, the numbers $m = m_1 + m_2$ and $n_1, n_2, n_3, n_4$ satisfy $(\dagger)$ and the associated constraints.

Further, the triangulation of the outside region $N_1$ determines a binary sequence of length $n_1 - 2$ as follows. Denote the exterior triangle in $N_1$ using the diagonal $A_0 A_{n_1}$ by $T_1$. If $n_1 \ge 3$, $T_1$ has a unique neighboring exterior triangle in $N_1$, denoted $T_2$. If $n_1 \ge 4$, the triangle $T_2$ has another neighbor in $N_1$ denoted $T_3$, etc. Thus we have a sequence of $n_1 - 1$ exterior triangles in $N_1$. We encode this sequence as follows. If $T_1$ uses the vertex $A_1$ as its third vertex we encode this by $0$ and if it uses $A_{n_1-1}$ we encode this by $1$. In each case there are two possible choices for the third vertex in $T_2$. If the one with smaller index is used we encode this by $0$ and if the one with larger index is used we encode this by $1$. Eventually, a sequence of $n_1 - 2$ $0$'s and $1$'s is constructed describing the choice of the third vertex in the triangles $T_1, \dots, T_{n_1-2}$. Finally, there is only one choice for the third vertex in the triangle $T_{n_1-1}$ (this triangle is uniquely determined by the previous one), so we get $2^{n_1-2}$ possible triangulations of $N_1$ encoded in a binary sequence of length $n_1 - 2$. Similarly, there are $2^{n_1-2}$ triangulations of the region $N_i$, $i = 1, 2, 3, 4$, encoded by binary sequences of length $n_i - 2$. Thus a binary sequence $w_N$ of length $n_1 - 2 + n_2 - 2 + n_3 - 2 + n_4 - 2 = n - m - 8$, uniquely determines the triangulations of the regions $N_1, N_2, N_3, N_4$ (once the regions are precisely determined within $P$, which is done once $m_1, m_2, n_1, n_2, n_3$ and $n_4$ are known).

It remains to uniquely encode the triangulation of the middle region $M$. Denote by $M_1$ the unique exterior triangle in $M$ using the diagonal $A_0A_{n_1+n_2}$. If $m \ge 2$, $M_1$ has a unique neighboring exterior triangle $M_2$ in $M$. If $m \ge 3$, the triangle $M_2$ has another neighbor in $M$ denoted $M_3$, etc. Thus we have a sequence of $m$ exterior triangles in $M$. We encode this sequence as follows. If $M_1$ uses the vertex $A_{n_1+n_2+1}$ as its third vertex we encode this by $0$ and if it uses $A_{n-1}$ we encode this by $1$. In each case there are two possible choices for the third vertex in $M_2$. If the one with smaller index is used we encode this by $0$ and if the one with larger index is used we encode this by $1$. Eventually, a sequence of $m$ $0$'s and $1$'s is constructed describing the choice of the third vertex in the triangles $M_1, \dots, M_m$. Thus a binary sequence $w_M$ of length $m$ uniquely determines the triangulation of the region $M$. In addition such a sequence $w_M$ uniquely determines $m_1$ and $m_2$ as the number of $0$'s and $1$'s respectively in $w_M$ and therefore also the exact position of the middle region $M$ within $P$ (once $n_1$ and $n_2$ are known), which in turn then exactly determines the position of all the regions considered in the figure.

The number of solutions of the equation $(\dagger)$ subject to the given constraints is equal to the number of positive integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = n - 3,$$
which is $\binom{n-4}{4}$ (a sequence of $n-3$ objects is split into $5$ nonempty groups by placing $4$ separators in the $n-4$ available positions between the objects). Thus the number of $7$-tuples $(m, n_1, n_2, n_3, n_4, w_M, w_N)$ describing triangulations as in the figure is
$$2^m \cdot 2^{n-m-8} \binom{n-4}{4} = 2^{n-8} \binom{n-4}{4}.$$

Finally, in order to get the total number of triangulations we multiply the above number by $n$ (since we could start building the triangulation at any vertex rather than at $A_0$) and divide by $2$ (since every triangulation is now counted twice, once as starting at one of the interior triangles and once as starting at the other).

Thus, the answer is
$$n2^{n-9} \binom{n-4}{4}$$