5. Label the $n$ vertices of a convex $n$-gon $S$ sequentially as $1,2, \cdots, n$. For any point outside $S$, the angle subtended by $S$ at this point is less than $180^{\circ}$. Within this angle, read out the numbers of each vertex in a clockwise direction, resulting in an $n$-term sequence. For example, for the convex pentagon shown in Figure 2, the sequence read from point $X$ is $(4,5,3,2,1)$, and the sequence read from point $Y$ is $(5,4,1,2,3)$. To ensure the uniqueness of the read sequence, the observation point should not be collinear with any two vertices of the convex $n$-gon $S$. For a convex $n$-gon $S$, let the total number of different $n$-term sequences that can be read be $f(S)$.
For all convex $n$-gons $S$, find the maximum possible value of $f(S)$.
(Lin Chang provided the problem)
5. The maximum value of $f(S)$ is $2\left(\mathrm{C}_{n}^{4}+\mathrm{C}_{n}^{2}\right)$.

For any two vertices $A, B$ of a convex $n$-gon $S$ and two observation points $X, Y$ outside $S$, when $X, Y$ are on the same side of the line $A B$, the order of $A, B$ read from $X, Y$ is the same; when $X, Y$ are on different sides of the line $A B$, the order of $A, B$ read from $X, Y$ is opposite. For example, for the vertices $A, B$ in Figure 5, the order read from point $X$ is $(A, B)$, and the order read from point $Y$ is $(B, A)$. Let the lines connecting all pairs of vertices of $S$ divide the exterior of $S$ into $m$ regions. Then, for any two points $X, Y$ in the same region, they are on the same side of any line $A B$, and the order of $A, B$ read from $X, Y$ is the same, so the sequences read from $X, Y$ are the same. For any two points $X, Y$ in different regions, there must exist a line $C D$ such that $X, Y$ are on different sides of the line $C D$, and the order of $C, D$ read from $X, Y$ is opposite, so the sequences read from $X, Y$ are different. Therefore, $f(S)=m$.

Connecting all pairs of vertices of the convex $n$-gon $S$, we get $t=\mathrm{C}_{n}^{2}$ lines, forming $2 t$ rays in the exterior of $S$, denoted as $L_{1}, L_{2}, \cdots, L_{2 t}$. Construct these rays sequentially. Let $L_{i}$ intersect the previous $i-1$ rays at $\alpha_{i}$ points, which divide $L_{i}$ into $\alpha_{i}+1$ segments, each segment dividing an existing region into two regions. Therefore, constructing $L_{i}$ increases the number of regions by $\alpha_{i}+1$. Thus,
$$
m=\sum_{i=1}^{2 t}\left(\alpha_{i}+1\right)=2 t+\alpha,
$$

where $\alpha=\sum_{i=1}^{2 t} \alpha_{i}$ is the number of intersection points of the $t$ lines in the exterior of $S$ (counting multiplicities, if a point is the common point of $k$ rays, it is counted $k-1$ times).

Each external intersection point is determined by two lines connecting vertices of $S$, i.e., by four vertices of $S$, and any four vertices can determine at most two external intersection points (as shown in Figure 6), so $\alpha \leqslant 2 \mathrm{C}_{n}^{4}$. Therefore,
$$
f(S)=m=2 t+\alpha \leqslant 2\left(\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{4}\right) .
$$

In the above formula, equality holds if and only if the $t$ lines connecting the vertices of the convex $n$-gon $S$ are pairwise non-parallel and have no three lines intersecting at the same point in the exterior of $S$. We can construct a convex $n$-gon that meets this condition by induction on $n$, thus achieving the maximum value of $f(S)$, which is $2\left(\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{4}\right)$.

When $n=3$, any triangle meets the requirement. Assume we have constructed a convex $n$-gon $S_{n}$ that meets the requirement, with the set of its vertices denoted as $D$, the set of lines connecting its vertices denoted as $T$, and the set of intersection points of $T$ in the exterior of $S_{n}$ denoted as $P$. Let the external region of the side $A B$ of $S_{n}$ be $F$. We prove that the union of $F$ and any point $C$ in $F$ (i.e., replacing $A B$ with edges $C A, C B$, as shown in Figure 7) is a convex $(n+1)$-gon $S_{n+1}$ that meets the requirement.

Draw a parallel line to each line in $T$ through each point in $D$. When $C$ is not on any of these parallel lines, the lines connecting $C$ to each point in $D$ are not parallel to any line in $T$, so the lines connecting the vertices of $S_{n+1}$ are pairwise non-parallel. Draw a line connecting each point in $D$ to each point in $P$. When $C$ is not on any of these lines, the lines connecting $C$ to each point in $D$ do not pass through any point in $P$, so the lines connecting the vertices of $S_{n+1}$ have no three lines intersecting at the same point in the exterior of $S_{n+1}$.

Since the number of parallel lines to each line in $T$ through each point in $D$ and the number of lines connecting each point in $D$ to each point in $P$ are both finite, there must exist a point $C$ in the region $F$ that is not on any of these lines (in fact, there are infinitely many such points), so $S_{n+1}$ meets the requirement.
In summary, the maximum possible value of $f(S)$ is $2\left(\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{4}\right)$.