# How many combinations of two diagonals are there that intersect outside the $n$-gon?

We have an $n$-gon with all diagonals in it and with no parallel diagonals and no three diagonals intersecting in a point.How many combinations of two diagonals are there that intersect outside the $n$-gon?

The answer given in the book is $\frac{n(n-3)(n-4)(n-5)}{12}$.Maybe $\frac{n(n-3)}{2}$ is the number of ways choosing a diagonal but when I calculate I see that if a diagonal is drawn both two other points should be onside the diagonal but I don't know how to calculate it.Any hints?

• In how many ways can you select a chord, then two extra vertices on an arc delimited by such a chord? – Jack D'Aurizio Apr 22 '17 at 15:36
• How much are you over-counting this way? This question is pretty straightforward. – Jack D'Aurizio Apr 22 '17 at 15:36

Select four distinct vertices of the polygon and consider their convex envelope $ABCD$.
This quadrilateral gives two couples of lines intersecting outside the polygon, $(AB,CD),(BC,AD)$.
In particular, by including the sides of the original polygon, there are $2\binom{n}{4}$ couple of lines intersecting outside. It is simple to finish from here.
• @TahaAkbari: sorry for before, the correct count is $$2\binom{n}{4}-n\binom{n-2}{2}+\frac{n(n-3)}{2}=\frac{n(n-3)(n-4)(n-5)}{12}.$$ – Jack D'Aurizio Apr 22 '17 at 22:32