# How many pairs of diagonals of of a odd sided regular polygon intersect within the interior the polygon?

How many pairs of diagonals of a $2n+1$ sided regular polygon intersect within the interior of the polygon?

(By interior I mean it shouldn't intersect on the vertex)

For a triangle there is no diagonal.

For a Pentagon five pair of diagonals intersect inside the interior of the Pentagon.

For a heptagon there are so many diagonals that I got confused.

Can someone help me out.

• This might be helpful --> wikihow.com/Find-How-Many-Diagonals-Are-in-a-Polygon Jul 3, 2016 at 3:33
• All of them intersect at least one other diagonal. What are you really asking? Jul 3, 2016 at 3:49
• Are you asking how many (unordered pairs) of diagonals intersect in the interior of the polygon? That is an interesting question with a suprisingly short answer. Jul 3, 2016 at 4:18
• @AndréNicolas Yes, I am exactly asking the same. I wrote in the question it intersect inside the polygon. Jul 3, 2016 at 6:44
• @RossMillikan Yes, they surely do intersect with sides, which is quite obvious. My question was how many pairs of diagonals intersect within the interior of the polygon. Jul 3, 2016 at 6:48

Choose $4$ vertices from the $2n+1$ available. Exactly one pair of diagonals determined by dividing these vertices into two pairs meet in the interior of the polygon.
Thus the required number of pairs of diagonals is $\binom{2n+1}{4}$.