Joining points 2

The number of configurations with zero intersection is the Catalan number $C_n.$ The number of configurations with one intersection is $\binom{2n}{n-2},$ as in your previous problem . The total number of configurations is $\frac{\binom{2n}2 \binom{2n-2}2 \cdots \binom{2}{2}}{n!} = \frac{(2n)!}{2^n n!},$ because the numerator of the left side is the number of ways to sequentially pick pairs of points to connect, and the denominator is the number of ways to pick a given set of pairs. So the answer to the question is $\frac{(2n)!}{2^n n!} - C_n - \binom{2n}{n-2},$ which for $n=5$ gives $945-42-120 = \fbox{783}.$