5. Let $n$ be a positive integer, and given $2 n$ distinct points on a circle. Divide these points into $n$ pairs, and connect each pair with a directed line segment. If any two directed line segments do not intersect, and there do not exist directed line segments $\overrightarrow{A B}$ and $\overrightarrow{C D}$ such that the convex quadrilateral with vertices $A, B, C, D$ in clockwise order, then the resulting figure is called "good". Find the number of good figures.
5. The number of good graphs is $\mathrm{C}_{2 n}^{n}$.

To prove this, label the points on the circumference in a counterclockwise direction as $a_{1}, a_{2}, \cdots, a_{2 n}$. Let $C$ be a good graph, and let $O(C)$ denote the set of starting points of all directed segments in this good graph.
We will prove this using mathematical induction:
Any $n$-element subset $S$ of the set $\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}$ is the unique preimage of a good graph.
For $n=1$, the case is easy to verify.
Assume that for a circle with $2(n-1)$ points, any $(n-1)$-element subset $S$ of the set $\left\{a_{1}, a_{2}, \cdots, a_{2 n-2}\right\}$ is the unique preimage of a good graph. Then, for a circle with $2 n$ points, consider any $n$-element subset $S$ of $\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}$. If there exists a good graph $C$ such that $S=O(C)$, then there must exist a $k$ $(k \in\{1,2, \cdots, 2 n\})$ such that
$a_{k} \in S$ but $a_{k+1} \notin S$ (with the convention that $a_{2 n+i}=a_{i}$). For $k+1<l<2 n+k$, according to the requirement that directed segments in a good graph do not intersect, the index $m$ of the directed segment $\overrightarrow{a_{m} a_{k+1}}$ with $a_{k+1}$ as the endpoint must satisfy $k+2<m<l-1$. This would form a convex quadrilateral in a clockwise direction, which is a contradiction, so $l=k+1$.

Conversely, if the graph $C$ contains the directed segment $\overrightarrow{a_{k} a_{k+1}}$, then this directed segment will not intersect with any other directed segments and will not form a convex quadrilateral in a clockwise direction with any other directed segments.

Therefore, by removing points $a_{k}$ and $a_{k+1}$ from $\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}$, and removing point $a_{k}$ from subset $S$, using the induction hypothesis, we know that selecting $n-1$ points from the remaining $2 n-2$ points corresponds uniquely to a good graph. This indicates that after adding points $a_{k}$ and $a_{k+1}$, the subset $S$ of $\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}$ uniquely corresponds to one good graph.
In summary, the number of good graphs is $\mathrm{C}_{2 n}^{n}$.