Let $n$ be positive integer and fix $2n$ distinct points on a circle. Determine the number of ways to connect the points with $n$ arrows (oriented line segments) such that all of the following conditions hold: [list] 	[*]each of the $2n$ points is a startpoint or endpoint 	of an arrow; 	[*]no two arrows intersect; and 	[*]there are no two arrows $\overrightarrow{AB}$ and 	$\overrightarrow{CD}$ such that $A$, $B$, $C$ and $D$ 	appear in clockwise order around the circle 	(not necessarily consecutively). [/list]
1. **Claim:** The number of ways to connect \(2n\) points with unoriented non-intersecting line segments is \(C_n\), the \(n\)th Catalan number.

2. **Proof:** Let \(f(n)\) denote the number of ways to connect \(2n\) points with unoriented non-intersecting line segments. We aim to prove that \(f(n) = C_n\).

3. **Base Case:** For \(n = 1\), there is only one way to connect 2 points with a single line segment, so \(f(1) = 1\).

4. **Recurrence Relation:** Consider a specific point \(A\). This point can be connected to \(n\) different vertices (since we must leave an even number of points on either side of our line). Thus, we have the recurrence relation:
   \[
   f(n+1) = \sum_{i = 0}^{n} f(i)f(n-i)
   \]
   This is known as Segner's recurrence relation, which is the defining recurrence for Catalan numbers. Therefore, \(f(n) = C_n\).

5. **Catalan Number Formula:** The \(n\)th Catalan number is given by:
   \[
   C_n = \frac{1}{n+1} \binom{2n}{n}
   \]

6. **Orientation of Arrows:** We need to prove that there are \(n+1\) ways to order the arrows. 

7. **Claim:** There must be at least one arrow between neighboring points.

8. **Proof by Induction:** 
   - **Base Case:** For \(n = 1\), the case is trivial as there is only one arrow.
   - **Inductive Step:** Assume the claim holds for \(n\). Consider a specific arrow. This arrow splits the circle into two cases we have already covered: one with \(n-x\) arrows and one with \(x-1\) arrows. Thus, our induction is complete.

9. **Arrangement of Arrows:** We now prove that there are \(n+1\) ways to arrange the arrows.
   - **Base Case:** For \(n = 1\), there is only one way to arrange the arrow.
   - **Inductive Step:** Consider the chord connecting two neighboring points. If this arrow goes counterclockwise, then we have exactly \(n\) cases. If this arrow goes clockwise, we need every other arrow to go counterclockwise, giving exactly 1 case. Adding these, we get \(n+1\) cases as desired.

\(\blacksquare\)

The final answer is \(\boxed{\binom{2n}{n}}\)