Joining points

So I get $\binom{2n}{n-2}$ for a general answer, but I get it in a weird way, by using Catalan numbers and then getting a bunch of cancellation. There must be some direct combinatorial argument, but I haven't thought too hard about it...

My argument starts by using the fact that the number of ways to draw chords with zero crossings is the $n$ th Catalan number. Then the idea is: pick a point $A,$ and then go around clockwise and pick the other three points $B,C,D$ that the crossing lines are going to hit (the segments will be $AC$ and $BD$ ). So we've got $2n$ choices for the point (say $A$ ), then we pick a number $2a$ of points in between $A$ and $B$ where we have $C_a$ ways to draw non-intersecting chords, then we've got the next point $B,$ then $2b$ more points, then $C,$ then $2c$ more points, then $D,$ then $2d$ more points. So the total number of ways is $\frac{2n}4 \sum_{a+b+c+d=n-2} C_aC_bC_cC_d,$ where we've divided by $4$ because we could have started at $B,C,$ or $D$ and gotten the same configuration.

Now the recursive formula $C_{m+1} = C_0C_m + C_1C_{m-1} + \cdots + C_mC_0$ for the Catalan numbers allows us to chop up the sum: if $a+b=k,$ $\begin{aligned} \frac{2n}4 \sum_{a+b+c+d=n-2} C_aC_bC_cC_d &= \frac{n}2 \sum_{k=0}^{n-2} \left( \sum_{a=0}^k C_a C_{k-a} \right) \left( \sum_{c=0}^{n-k-2} C_c C_{n-k-2-c} \right) \\ &= \frac{n}2 \sum_{k=0}^{n-2} C_{k+1} C_{n-k-1} \\ &= \frac{n}2 \sum_{j=1}^{n-1} C_jC_{n-j} \\ &= \frac{n}2 \left( -C_0C_n + -C_nC_0 + \sum_{j=0}^n C_j C_{n-j} \right) \\ &= \frac{n}2 (C_{n+1}-2C_n). \end{aligned}$

The explicit formula $C_n = \frac1{n+1} \binom{2n}{n}$ plus some algebra lets us simplify that expression to $\binom{2n}{n-2}.$ For $n=5,$ this is $\binom{10}3 = \fbox{120}.$