3 deleted 13 characters in body

If you let me do something silly, then I can put all the points in a line. Technically this gives an infinite number of points of intersection, since every (non-vertex) point on the line segment is a point of intersection of two edges. If instead I count the number of pairs of edges which contain a non-vertex in their intersection then I still get

\begin{equation*} \binom{\binom{n}{2}}{2} - \sum_{1 < i \leq j < n} (i-1)(n-j), \end{equation*}

which I guess is maximum (it's more than $\binom{n}{4}$).

2 added 3 characters in body

If you let me do something silly, then I can put all the points in a line. Technically this gives an infinite number of points of intersection, since every (non-vertex) point on the line segment is a point of intersection of two edges. If instead I count the number of pairs of edges which contain a non-vertex in their intersection then I still get

\begin{equation*} \binom{\binom{n}{2}}{2} - \sum_{1 < i < \leq j < n} (i-1)(n-j), \end{equation*}

which I guess is maximum (it's more than $\binom{n}{4}$).

1

If you let me do something silly, then I can put all the points in a line. Technically this gives an infinite number of points of intersection, since every (non-vertex) point on the line segment is a point of intersection of two edges. If instead I count the number of pairs of edges which contain a non-vertex in their intersection then I still get

\begin{equation*} \binom{\binom{n}{2}}{2} - \sum_{1 < i < j < n} (i-1)(n-j), \end{equation*}

which I guess is maximum (it's more than $\binom{n}{4}$).