The answer is ${2n\choose n}$. Select the starting points of all arrows. Now it is easy to see that it uniquely determines the orientantion for each arrow.

You can also derive a straightforward Catalan-like recurrence. After that, bash out small cases, and strong induct. The recurrence can be solved with simple genfun. Denote $f(n)$ as the answer for $n$. We will show that $f(n)=\binom{2n}{n}$. We can easily show (isw catalan) that $$f(n)=2 \cdot \sum_{i=0}^{n-1} C_{n-i-1}f(i)$$ Now after strong induction hypothesis, all we need to show is that $$\binom{2k+2}{k+1} = 2 \sum_{i=0}^k C_{k-i} \cdot \binom{2i}{i} $$ Set $G(x)= \sum_{i=0}^k C_i x^{i+1}$, and we will calculate the coefficient of $x^{k+1}$ of $2G(x)G'(x)$ in two ways. Method 1. $G'(x) = \sum_{i=0}^k \binom{2i}{i} x^i$, so the coefficient of $x^{k+1}$ is $2 \sum_{i=0}^k C_{k-i} \cdot \binom{2i}{i} $. Method 2. The coefficient of $x^{k+2}$ in $G(x)^2$ is $C_{k+1}$ by Catalan Numbers' recurrence, so the coefficient of $x^{k+1}$ in $(G(x)^2)' = 2G(x)G'(x)$ is $(k+2) C_{k+1} = \binom{2k+2}{k+1}$. Therefore, we have our desired equality and we are done by induction.

Wolowizard wrote: The answer is ${2n\choose n}$. Select the starting points of all arrows. Now it is easy to see that it uniquely determines the orientantion for each arrow. It may violate the third condition

rkm0959 wrote: You can also derive a straightforward Catalan-like recurrence. After that, bash out small cases, and strong induct. The recurrence can be solved with simple genfun. @below There's a unique way for any choice of n start points to draw the arrows so that it follows all conditions. Take hexagon ABCDEF as an example, if A, D and E are chosen, there is no way

A -> F E-> B D->C

If we define the answer $d_n$, and $c_n$ the $n$th catalan number, we could got $d_n$=$2(c_0 d_{n-1}+c_1 d_{n-2}+... c_{n-1}d_0)$ And using strong induction, the answer is $(n+1)$$c_n$=${2n\choose n}$

rkm0959 wrote: A -> F E-> B D->C oops, counter-clockwise is eligible, sorry my bad

The answer is $\binom{2n}{n}$, which can be guessed from the first few cases ($n \le 3$). There are three classes of approaches I'm aware of, which I will outline below (and leave others to fill in the details). Show that a configuration is uniquely determenid by the choice of which $n$ points are endpoints and which $n$ points are startpoints. Show that given an unoriented connection (which are counted by the Catalan numbers $C_n = \tfrac{1}{n+1} \tbinom{2n}{n}$), there are exactly $n+1$ ways to orient it. The Korean deputy Sang-Il Oum gives an approach to this proof by considering the dual graph of the configuration in the plane; this gives a tree with $n+1$ vertices, and selecting a source (root) of the graph (thus directing the tree) turns out to induce an orientation of the arrows. Directly by recursion and generating functions. Indeed, if $D_n$ is the answer to the question and $C_n = \tfrac{1}{n+1}\tbinom{2n}{n}$ is the Catalan number then one can prove that \[ D_n = 2 \sum_{k=1}^n D_{n-k} C_{k-1}. \]The corresponding generating functions $d(x) = \sum_n D_n x^n$ and $c(x) = \sum_n C_n x^n = (2x)^{-1}(1-\sqrt{1-4x})$ thus satisfy $d(x) = 2xc(x)d(x)+1$, hence \[ d(x) = \frac{1}{1-2xc(x)} = (1-4x)^{-1/2} = \sum_n \binom{2n}{n} x^n. \] Remarkably, combining the first and second approach gives a proof of the Catalan number formula.

My solution. For the sake of convenience we let the $2n$ points be the vertices of a regular polygon, as the conditions aren't changed by this. We start with putting down non-oriented edges that the first two conditions are satisfied. We get $\frac{1}{n+1} \cdot \binom{2n}{n}$ ways to do it. Pick up any such configuration. I claim that there are exactly $n+1$ ways to satisfy the third condition. We establish this by induction. Base case of $n=1$ is trivial. Observe that at least one pair of two consecutive vertices is joined by an edge. This is true because there must exist a smallest edge and no edge from one 'region' to the other 'region' of the circle exists, where an edge divides the circle's interior into two regions. Call this edge $e$. This edge can be oriented clockwise or anticlockwise. If it is anticlockwise, we use the case for $n-1$, and we have $n$ ways to orient the other edges. If it is clockwise, there is only one orientation that is possible. It is automatic to see that when we start moving around the circle in an anticlockwise sense from the vertices at the ends of edge $e$, the first encountered vertex of an edge must be at the tail end of the arrow. This proves the claim. Hence the required number is $\binom{2n}{n}$.

WLOG the $2n$-gon is regular. We claim that for any undirected pairing, there are $n+1$ ways to orient the segments so the condition is satisfied so it follows that the answer is $\binom{2n}n$. Define the length of a directed segment between $A$ and $B$ as the number of vertices of the $2n$-gon that are passed when walking from $A$ to $B$ clockwise around the circle counting exactly one of $A$ and $B$. The key observation is that if some directed segment $A\rightarrow B$ has minimal length which is at most $n$, then the rest of the directions are uniquely determined. Indeed, all segments along the walk from $B$ to $A$ clockwise around the circle must be directed against the walk, or we would contradict the condition. Furthermore, all segments along the walk from $A$ to $B$ clockwise around the circle must be directed against the walk, or we would find a directed segment of smaller length, a contradiction. Hence, given this minimal segment the rest of the directions are uniquely determined. Now, note that every segment other than a diameter of the circumcircle of the $2n$-gon has exactly one orientation with length at most $n$, while either orientation of a diameter has length $n$. If our undirected pairing has no diameters, then we may choose none or exactly one segment to have minimal length at most $n$. If we do have a diameter, then we cannot choose none of the segments to have minimal length at most $n$, but we can orient the diameter either way to become the minimal segment. In either case, we have exactly $n+1$ ways to direct the segments, as desired.

@below fixed, thanks.

@above We can have more than one clockwise segment, consider $P_1\rightarrow P_4$ and $P_2\rightarrow P_3$ for clockwise $P_1P_2P_3P_4\ldots P_{2n}$ with $n$ large.

We can also solve the recursion in v_Enhance solution without generating functions: \[ D_n = 2 \sum_{k=1}^n D_{n-k} C_{k-1} \Leftrightarrow D_{n+1}=2 \sum_{k=0}^{k=n} D_{n-k} C_{k} \]Where $C_{k}$ is the $k^{th}$ Catalan number. We proceed by induction. Assume $D_{k}=\binom{2n}{n}$ for all $k \leq n$. We will count in two ways the number of paths on an $(n+1) \text{x} (n+1)$ board going left and up from the bottom-left corner to the top-right corner. By stars and bars this is $\binom{2(n+1)}{n+1}$. We set the coordinates of the bottom-left corner to be $(-1,-1)$ and the top-right corner $(n,n)$ First we consider if the first move is $up$. The case for if this move is $right$ will be similar but a reflection across the main diagonal. Consider the first time this path hits the main diagonal at $(k,k)$ where $0 \leq k \leq n$. The move going to this square must have been a right so we go $(k-1,k) \rightarrow (k,k)$. The $2k$ moves from $(-1,0) \rightarrow (k-1,k)$ must have stayed above the main diagonal shifted up by one (i.e. the line $(-1,0),(0,1), \cdots, (n-1,n)$) and it is well-known the number of ways to do this is $C_k$. The $2(n-k)$ moves from $(k,k) \rightarrow (n,n)$ are freely chosen so by stars and bars this is $\binom{2(n-k)}{n-k}=D_{n-k}$. This yields a total number of paths starting with an $up$ to be: $\sum_{k=0}^{k=n} C_k D_{n-k}$ and the process with where the first move is $right$ is the same but instead, we consider staying below the main diagonal. Hence the total number of paths is: \[2\sum_{k=0}^{k=n} C_k D_{n-k}\]But using the recurrence and the two ways of counting we get: \[D_{n+1}=2\sum_{k=0}^{k=n} C_k D_{n-k}=\binom{2(n+1)}{n+1}\]So we're done by induction.

Oops, didn't notice the clever solutions. However, this still feels like a glorified AIME problem. Let $f(n)$ be the answer to the problem. We use the standard Catalan recurrence idea. Basically, choose some vertex to be a starting point, and suppose the end is such that on one side of the arrow there are $2k$ points, and on the other there are $2(n-k-1)$. Furthermore, if the arrow is pointing "down", choose the $2k$ side to be on the "left". This ensures that the number of ways to fill the left side is $f(k)$ (since arrows can point either way and not violate condition wrt to the chosen arrow), and the number of ways for the right side is $C_{n-k-1}$ (since all the arrows have to point "down"). Therefore, we have that \[\frac{1}{2}f(n)=\sum_{k=0}^{n-1}C_{n-k-1}f(k),\]since our chose arrow's direction can be flipped. Let us now compute the generating function of either side, and to do so, let $F(x)=\sum_{n\ge 0}f(n)x^n$. Our goal is to find $F(x)$. Note that the GF of the left side is $\frac{1}{2}(F(x)-f(0))=\frac{1}{2}(F(x)-1))$ (we'll sum each side times $x^n$ from $n=1$ to $\infty$). The GF of the right side is \begin{align*} \sum_{n\ge 0}\sum_{k=0}^{n-1}C_{n-k-1}f(k)x^n &= \sum_{k\ge 0}f(k)x^k\cdot x\sum_{n\ge k+1}C_{n-k-1}x^{n-k-1} \\ &= xF(x)\left(\sum_{m\ge 0}C_m x^m\right) \\ &= F(x)\cdot\frac{1-\sqrt{1-4x}}{2}. \end{align*}Therefore, \[F(x)-1=(1-\sqrt{1-4x})F(x),\]so $F(x)=\frac{1}{\sqrt{1-4x}}$. This implies that $f(n)=\boxed{\binom{2n}{n}}$.

Slightly different finish than above generating function posts. Let $f(n)$ be the answer. As many other users have noted already, we get the recurrence \[ f(n) = 2\sum_{k=0}^{n-1} \frac{1}{k+1} \binom{2k}{k} f(n-1-k). \]We claim that $f(n) = \boxed{\binom{2n}{n}}$. Induct on $n$, the base case of $n=1$ trivial. We want to show that \[ \binom{2n+2}{n+1} = 2\sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k} f(n-k) = \sum_{k=0}^n \frac{2}{k+1} \binom{2k}{k} \binom{2(n-k)}{n-k}. \]Write \begin{align*} &~ \sum_{n\ge 0} \left[ \sum_{k\ge 0} \frac{2}{k+1} \binom{2k}{k} \binom{2n-2k}{n-k} \right] x^n =\sum_{k\ge 0} \frac{2}{k+1} \binom{2k}{k} \sum_{n\ge 0} \binom{2n-2k}{n-k} x^n \\ &= \sum_{k\ge 0} \frac{2}{k+1} \binom{2k}{k} x^k \cdot \frac{1}{\sqrt{1-4x}} = \frac{1}{\sqrt{1-4x}} \cdot 2 \cdot \frac{1-\sqrt{1-4x}}{2x} \\ &= \frac{1-\sqrt{1-4x}}{x\sqrt{1-4x}} = -\frac{1}{x} + \frac{1}{x\sqrt{1-4x}} = \frac{1}{x} \left( -1 + \frac{1}{\sqrt{1-4x}} \right) \\ &= \frac{1}{x} \left(-1 + \sum_{n\ge 0} \binom{2n}{n} x^n \right) = \frac{1}{x} \sum_{n\ge 0} \binom{2(n+1)}{n+1} x^{n+1} \qquad \text{(just shifting the index)} \\ &= \sum_{n\ge 0} \binom{2n+2}{n+1} x^n. \end{align*}This completes the induction.

Define an arrow starting from $A$ and ending at $B$ as "clock" if the minor arc $\overarc{AB}$ is going in the clockwise direction. Then, take any configuration with $n$ lines (not arrows) such that no two lines intersect and each point is on exactly one line. Then, when we turn these lines into arrows, at most $1$ of these lines can be clock. This is because if there were two clock arrows, then it would contradict our third criterion. There are a total of $n+1$ ways to choose a clock line (we could also choose $0$ lines to be clock). For each line that we choose to be clock, we will choose every other line to be not clock, which clearly satisfies the problem statement. Therefore, there are $(n+1)$ times the total number of ways to choose the lines. It is well known that the total number of ways to choose the lines is: $$\frac{1}{n+1}\binom{2n}{n}$$so our answer is $\binom{2n}{n}$.

We claim that the answer is $\binom{2n}{n}$, which we will show with Catalan winduct. Consider a ray $\overrightarrow{AB}$. Now consider the clockwise arc $AB$; we can place whatever the hell we want here and it'll be fine(from the reference point of $\overrightarrow{AB}$. Now, consider the counterclockwise arc $AB$. Everything on this half must be counterclockwise. WLOG orient to focus on the ray starting at some vertex $A$ and ending at $B$. Let's say the counterclockwise region has $2k$ points, and the clockwise region has $2n - 2k - 2$ points. For the counterclockwise arc half, there's no assigning directions anymore. We can just focus on the number of ways to partition the $2k$ points there into disjoint nonintersecting segments. This is a very well known problem, and it is the $k$th Catalan number given by $C_k = \frac{1}{k + 1} \binom{2k}{k}$. The clockwise arc half is simply $F(n - k - 1)$, where we let $F(n)$ be the number of ways to make such a partition. However, we must multiply by $2$, since analogous logic holds for whenever our ray goes into $A$. Basically, everything I said above gets clockwise and counterclockwise swapped. Thus we get the recursion \begin{align*} F(n) = 2 \sum_{i = 0}^{n - 1} C_i \cdot F(n - i - 1). \end{align*}To solve this, our base cases are $F(1) = 2$, $F(2) = 6$ which satisfy the answer. Now, we just need to genfunc \begin{align*} 2 \sum_{i = 0}^{n - 1} \frac{1}{i + 1} \binom{2i}{i} \cdot \binom{2n - 2i - 2}{n - i - 1}. \end{align*} Now for a bunch of algmanip: \begin{align*} F(n) &= \sum_{n = 0}^{\infty} \sum_{i = 0}^{n - 1} \frac{2}{i + 1} \binom{2i}{i} \binom{2n - 2i - 2}{n - i - 1} x^n \\ &= \sum_{i = 0}^{\infty} \sum_{n = i + 1}^{\infty} \frac{2}{i + 1} \binom{2i}{i} \binom{2n - 2i - 2}{n - i - 1} x^n \\ &= 2 \sum_{i = 0}^{\infty} C_i \sum_{k = 0}^{\infty} \binom{2k}{k} x^{i + k + 1} \\ &= 2x \sum_{i = 0}^{\infty} C_i \cdot \frac{x^i}{\sqrt{1 - 4x}} \\ &= \frac{2x}{\sqrt{1 - 4x}} \sum_{i = 0}^{\infty} C_i x_i \\ &= \frac{1}{\sqrt{1 - 4x}} - 1 \\ &= 0x^0 + \binom{2}{1}x + \binom{4}{2}x^2 + \cdots, \end{align*}so we can extract our desired answer as $\binom{2n}{n}$.

The answer is 2nCn. We first treat the arrows as segments and ignore their directions; it's well-known that the number of ways to choose such segments is the nth Catalan number, equal to (2nCn)/(n+1). Now, we can make either 0 or 1 segments a clockwise-pointing arrow and everything else counterclockwise, which can be done in n+1 ways, giving us the answer upon multiplication. Sent from Montgomery Blair HS

Straightforward for a RMM5, once you find the answer. I claim the answer is ${2n \choose n}$. We proceed strong inductively. Let the number of possible connections for $2i$ points be $a_i$. First, observe that the number of undirected connections for $2n$ points is the $n$th Catalan number $C_n$ by definition. Then, upon connecting an arrow, the points on one side of the arrow can be connected without attention to the orientation of that arrow, and the points on the other side have their arrow orientation uniquely fixed. As a result, we have the recursive formula $$a_n = 2\sum_{i \geq 0 } a_i C_{n-1-i},$$and thus it suffices to show that $${2n \choose n} = 2\sum_{i \geq 0} {2i \choose i} C_{n-2-i}.$$Proceed with Snake Oil. Observe that \begin{align*} \sum_{i \geq 0} {2i \choose i} \sum_{n \geq 0} C_{n-1-i} x^n &= \sum_{i \geq 0} {2i \choose i} x^{i+1} \cdot \frac{1-\sqrt{1-4x}}{2x} \\ &= (x)\left(\frac 1{\sqrt{1-4x}}\right)\left(\frac{1-\sqrt{1-4x}}{2x}\right) &=\frac 1{2\sqrt{1-4x}} + \frac 12. \end{align*}Notice that the coefficient of $x^n$ here is $\frac 12 {2n \choose n}$, so multiplying by the factor of 2 finishes.

Loved this problem! Let $c_n$ denote the ways to choose $n$ line segments each with distinct endpoints (not arrows) such that no two intersect. Notice that \[c_n=c_0c_{n-1}+c_1c_{n-2}+\dots+c_{n-1}c_0\]which can be proven by considering the segment with endpoint as a fixed point, and $c_0=1,c_1=1$. These are the exact criterion for the catalan number recurrence, hence $c_n=\binom{2n}{n}\frac{1}{n+1},$ the $n$'th catalan number. Now, go to the problem at hand; let $t_n$ denote the answer for $2n$ points on the circle. Then, considering a fixed endpoint of an arrow, we find \[t_n=2\sum_{i=0}^{n-1}c_it_{n-i-1}\]where the factor of $2$ comes from the two arrow orientations and each term in the sum is a consequence of conditions 2 and 3. Next, consider the generating function \[A(x)=\sum_{n\geq 0}t_nx^n=\sum_{n\geq 0}2x^n\sum_{i=0}^{n-1}c_it_{n-i-1}.\]This inner sum is the coefficient of $x^n$ in \[\left(\sum_{i\geq 0}x^it_i\right)\left(\sum_{j\geq 0}c_jx^j\right)=\left(\sum_{i\geq 0}x^ic_i\right)A(x).\]We find that \[A(x)=1+2x(A(x))\left(\sum_{i\geq 0}x^ic_i\right)=1+2x\cdot\frac{1-\sqrt{1-4x}}{2x}A(x)=1+\left(1-\sqrt{1-4x}\right)A(x)\]where the $1$ comes from $x^0t_0=1.$ We solve for A(x) to get \[A(x)=\frac{1}{\sqrt{1-4x}}=\sum_{i \geq 0} \binom{2i}{i}x^i.\]The answer is $\boxed{\binom{2n}{n}}.$

Claim 1 The number of ways to "cut" a polygon with $2n$ vertices with $n$ non-intersecting segments is: \[\frac{1}{n+1}\binom{2n}{n}=C_n\] We fix some vertex and see all the possible segments for this vertex. Let the distance of such segments is the number of vertices counterclockwise from the set vertex to the endpoint. So if the number of ways is $f(n)$ we see that we divide the polygon on two parts and they are independent one another, so \[f(n)=\sum_{i=0}^{n-1}f(i)\cdot f(n-1-i)\] Where $f(0)=1$. This is the recursion formula of the Catalan numbers, so we can easily prove $f(n)=C_n$. Claim 2 For every "cutting" with segments of the polygon we have $n+1$ ways to direct the segments so that the three conditions hold We will prove it with induction by the number of vertices. Obviously we have a segment between to neighbouring vertices. Now we let it be $\overrightarrow{AB}$. If $\overrightarrow{AB}$ is clockwise, then all the others segments have a single way to be directed. If $\overrightarrow{AB}$ is anticlockwise, then there is no segment $\overrightarrow{CD}$ such that the $A$, $B$, $C$ and $D$ to be clockwise order. So the ways to direct the segments is the same as for $2n-2$ vertices, which is $n$ That means for every "cutting" we have $n+1$ ways to direct them. So all the ways to make the directed segemnts are \[C_n\cdot{(n+1)}=\binom{2n}{n}\]

Pretty easy to see how to do this problem We know the number of triangulations in a polygon is $\frac1{n+1}\binom{2n}n$. Now, we know there exists a segment AB connecting adjacent vertices; if AB is clockwise, the rest are uniquely determined to be counterclockwise. On the other hand, if AB is counterclockwise, AB will never violate the counterclockwise condition. Now, we use induction. The base case n=2,2n-2=2 has 2=n ways to orient the other arrow; let the inductive hypothesis be that there are n ways. In our AB counterclockwise case, this reduces to 2n-2 vertices, which by inductive hypothesis has n ways, so there are indeed now n+1 ways to make arrows for 2n vertices, which holds the induction since n+1=2(n+1)-2. Now, the answer is the triangulations times n+1, so our answer is indeed $\binom{2n}n$. $\blacksquare$

We claim the answer is $\binom{2n}{n}$. Let the desired number on $2n$ vertices be $A_n$. Now we define another sequence $C_1, C_2, \dots$ that denotes the number of non-directed arrows on the $2n$ vertices satisfying the first two conditions. This is well known to be the Catalan numbers. Use induction. for $n = 1$ there are 2 ways, as desired. Position the $2n$ vertices so that there exists a unique vertex that has $y$-coordinate the greatest and it is vertically symmetrical. First assume that the arrow on this vertex points away. It is easy to see that the arrow must partition the vertices into two sections of even number of vertices (possibly of size 0). Let the sizes by $2k, 2n-2-k$ with the $2k$ vertices on the right. They are not affected by the arrow, so the number of ways we can connect these points is just $A_k$. For the $2n-2-2k$ other vertices, note that if we pair the points up with un-directed arrows, the arrows have exactly one direction they can be in (all pointing down). Therefore there are exactly $C_{n-1-k}$ ways to connect these points. (with $C_0 = 1$) If the arrow is pointing towards the topmost vertex, we obviously have the exact same number of ways by symmetry. It now suffices to prove that $\displaystyle \binom{2n+2}{n+1} = 2 \sum_{i = 0}^{n} C_{n-i}\binom{2i}{i}$. Using the formula $C_{n+1} = C_0C_n+C_1C_{n-1}+\dots + C_nC_0$, we have \begin{align*} (n+2)C_{n+1} &= (n+2)(C_0C_n+C_1C_{n-1}+\dots+C_nC_0) \\ &= \left((n+1)C_0C_n+nC_1C_{n-1}+\dots C_nC_0\right) + \left(C_0C_n+2C_1C_{n-1}+\dots + (n+1)C_nC_0\right) \\ &= 2\sum_{i=1}^n C_{n-i}\binom{2i}{i}, \end{align*}as desired.

We claim that the answer is $\boxed{\binom{2n}{n}}$. Claim: The number of ways to connect $2n$ points with unoriented non intersecting line segments is $C_n$ or the $n$th Catalan number. proof: Let $f$ denote the number of ways to connect $2n$ points with unoriented non intersecting line segments. We try to prove that $f(n) = C_n$. Notice that $f(1 = 1$. Notice also that if we consider a specific point $A$. This point can go to $n$ different vertices (we must leave an even number of dots on either side of our line). Thus we can see that we have the recurrence relation that \[f(n+1) = \sum_{i = 0}^{n} f(i)f(n-i)\]Notice that this is just Segner's recurrence relation. Thus, $f(n) = C_n$. $\square$ Now since $C_n = \frac{1}{n+1} \binom{2n}{n}$, we simply need to prove that there are $n+1$ ways to order the arrows. Notice that Claim: There must be at least one arrow between neighbors. proof: We prove the result by induction. Our base case is trivial Now for the inductive step, we take a specific arrow. Notice that this arrow splits the circle into two cases we have already covered: one with $n-x$ arrows and $x-1$ arrows. Thus our induction is complete. $\square$ We now try to prove that there are $n+1$ ways to arrange the arrows. We prove the result by induction. Our base case of $1$ is quite obvious. Now for the inductive step. Consider the chord connecting two neighboring points. We can see that if this arrow goes counterclockwise then we have exactly $n$ cases. However, if this arrow goes clockwise we need every other arrow to go counterclockwise giving exactly $1$ cases. Adding we get $n+1$ cases as desired. $\blacksquare$

Let $a_n$ be the numbers of ways to connect the points with $n$ arrows satisfying the problem condition. We will claim that $a_n = \binom{2n}{n}$. Let $b_n$ be the number of ways to connect $n$ segment such that each of the $2n$ is a startpoint or endpoint of an segment and no two segments intersect. The recurrence relation of sequence $(b_n)_{n \ge 1}$ is $b_n = \sum_{k=1}^{n} b_{i-1}b_{n-i}$, so it's clear that $b_n$ is the $n$th Catalan number, so $b_n = \frac{1}{n+1}\binom{2n}{n}$. Therefore, we only need to prove that $a_n = (n+1)b_n$ for all $n \ge 1$. Let $P$ be one of the points on the circle. The arrow including $P$ divides the circle into two arcs. Let $2k, 2n - 2 - 2k$ be the numbers of points on the two arc that we mentioned before. Regarding the direction of the arrow including $P$, the number of ways to put direction on the segments is $b_ka_{n-k-1}$ or $b_{n-k-1}a_k$. Therefore $a_n = \sum_{k=0}^{n-1} b_ka_{n-k-1} + b_{n-k-1}a_k$. We now prove that $a_n = (n+1)b_n$ by induction. The base case is clear. Now assume $a_i = (i+1)b_i$ for $1 \le i \le n-1$. Then $a_n = \sum_{k=0}^{n-1} b_ka_{n-k-1} + b_{n-k-1}a_k = \sum_{k=0}^{n-1} (n+1)b_kb_{n-k-1} = (n+1)b_n$, thus $a_n = (n+1)b_n = \binom{2n}{n}$. Hence, we're done. $\blacksquare$

The answer is $${2n\choose n}=(n+1)C_n.$$ Since it is well known that there are $C_n$ non-intersecting matchings, we will show that given any matching, there are exactly $n+1$ ways to orient the segments. We will use induction. This is clearly true for $n=1$. Now, note that any matching must have some edge between two consecutive vertices (e.g. it obviously is true for $n=1$, for larger $n$ each edge in the matching breaks it apart into smaller $n$ and we induct down). We call the orientation of this edge problematic if it goes in the direction such that going clockwise is the arc containing all other vertices. If this edge is oriented problematically, then the orientations of all other edges are forced, as one of the directions will contradict the no clockwise cycle condition, giving $1$ solution. If it is not oriented problematically, then the two vertices can effectively be removed from the problem, as the arc it constrains contains no points, so this case has $n$ ways by induction as we are assuming $n-1$ has $n$ ways. Thus, there are $n+1$ ways to orient the edges, as desired.

We claim that the answer is $\binom{2n}{n}$. To prove this, we will show that choosing the $n$ startpoints and $n$ endpoints uniquely determines the entire configuration. Consider some startpoint $A$ such that its counterclockwise neighbor $B$ is an endpoint. Such a point must always exist; just choose a startpoint and move counterclockwise until an endpoint is reached. We can see that $\overrightarrow{AB}$ must be an arrow. Otherwise, the arrow from $A$ and the arrow to $B$ will either intersect or form a clockwise loop, both of which are illegal. The arrow $\overrightarrow{AB}$ can never be in a clockwise loop, so we can pretend the two points no longer exist. We can repeat this process, choosing another startpoint with a counterclockwise neighbor that is an endpoint, which are then forced to form an arrow. This continues until we only have two points left, when there is only one option for the final arrow. Thus, choosing the $n$ startpoints forces exactly one final configuration, so there are exactly $\binom{2n}{n}$ ways, as claimed.

It is well-known that if the arrows were instead undirected segments, then there are $C_n$ possible ways, where $C_n$ is the $n$th catalan number. Now, we claim that from any given pairing, there are exactly $n+1$ ways to pick the direction of each segment. $\textit{Lemma: }$ In the current pairing, there is at least one pair of adjacent points that are connected by a segment. We show this by strong induction. The base case, $n=1,$ is trivial. Suppose that our proposition holds for $n \leq k$ for some positive integer $k.$ Then consider $2k+2$ points on a circle. No matter what two points we pair, clearly we will be left with $2$ cases with $n \leq k.$ Therefore, by our inductive hypothesis there is at least one adjacent connected pair. QED Now, we show our claim with induction. The base case, $n=1,$ is trivial. Suppose that the claim holds for $n = k$ for some positive integer $k.$ Now, suppose that $n=k+1,$ and consider one of the adjacent connected pairs. Then if the arrow points clockwise then every other segment has a fixed direction, giving $1$ possibility. If the arrow point counterclockwise, we can simply ignore this arrow and we are reduced to the case with $n=k,$ hence there are $n+1$ possibilities in this case, and adding gives $n+2$ possibilities in total, so our induction is done. QED Therefore, there are $\displaystyle (n+1)C_n = \binom{2n}{n}$ ways to pick a configuration.

Label the points $a_1, a_2, \ldots, a_{2n}$, and denote $f(n)$ as our desired answer and $C_n$ as the $n$th Catalan number. First notice that two connected vertices must have different parity indices. Suppose an arrow connects $a_1a_{2k}$, and WLOG points towards $a_{2k}$. Then the orientation of the arrows connecting the $2k-2$ points along the clockwise path from $a_1$ to $a_{2k}$ are independent of $\overrightarrow{a_1a_{2k}}$, while the arrows connecting the $2n-2k$ on the other side are fixed. Thus we have the recurrence \[f(n) = \sum_{k=1}^n 2 f(k-1) C_{n-k},\] from which we can prove using base cases and generating functions that our solution is \[f(n) = \boxed{\binom{2n}{n}}.\]

Notice that if we dont order the arrows, there are $C_n=\frac1{n+1}\binom{2n}{n}$ ways to choose the chords so that none intersect. Its not so hard to see that any set of non intersecting chords can be given a direction in exactly $n+1$ ways. This gives us the answer of $\binom{2n}{n}$

The answer is ${2n\choose n}.$ We proceed by induction; we first count the number of non-oriented configurations, call that number $F(n)$. The idea is to fix a segement and incuct down to get : \[\displaystyle F(n)=\sum_{i=0}^{n-1} F(i)F(n-1-i)\]Which has the same formula as the Catalan numbers. Thus, it is easy to prove that $\displaystyle F(n)=\frac{{2n\choose n.}}{n+1}.$ Now, it suffices to prove that the number of possible orientations of the segments is exactly $n+1.$ Again, this is done by induction by looking at a segement between consecutive dots. If it's clockwise-oriented, then the configuration is forced. Otherwise, that edge doesn't matter, so we can delete it and induct to orient the remaining segments; there are $n$ ways to do so. Hence the number of orientations is exactly $1+n=n+1.$