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?

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

  • $\begingroup$ But when we have to vertexes selected next to each other we can have one such intersection. $\endgroup$ – Taha Akbari Apr 22 '17 at 15:59
  • $\begingroup$ But it doesn't gives the answer wanted:wolframalpha.com/input/… $\endgroup$ – Taha Akbari Apr 22 '17 at 18:29
  • $\begingroup$ @TahaAkbari: sorry for before, the correct count is $$2\binom{n}{4}-n\binom{n-2}{2}+\frac{n(n-3)}{2}=\frac{n(n-3)(n-4)(n-5)}{12}.$$ $\endgroup$ – Jack D'Aurizio Apr 22 '17 at 22:32

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