"incidental" intersections of a complete graph in the plane Given a complete graph of n vertices (no three of which are no collinear) in the plane and straight edges, what is the maximal possible number of "incidental intersections" of edges, i.e., number of non-vertices at which two distinct edges intersect each other, not counting multiplicity?
This is a question that I pose to the students in my Mathematics for Elementary School Teachers as a way to understand mathematical conjecturing and proving -- and not always finding the solution.  But it occurs to me that it might be handy to know whether the answer is actually known or not.  
 A: Assuming straight line segment edges, any 4 vertices determine 6 edges with at most one non-vertex intersection, so you can't have more than $n$-choose-4. You will have $n$-choose-4 if no vertex is interior to any triangle of vertices, which is to say if all $n$ vertices lie on the boundary. 
A: This problem (although phrased slightly differently) is #1.3.5 in Loren C. Larson's "Problem Solving Through Problems."
The rephrase of this is, "On a circle, n points are selected and the chords joining them in pairs are drawn.  Assuming no three of the chords are concurrent (except at the endpoints), how many points of intersection are there?"
This is $\binom{n}{4}$, as any four of our points on the circle determine a (unique) intersection point, and any intersection point determines the boundary chords.
[This might've been better as a comment on Gerry's answer, but there's a whole family of similar problems to the question asker's in that chapter of the book (which are also excellent problems to give to a math club.)]
A: 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 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 $\binom{n}{4}$).  
