Counting the triangles formed by the sides and diagonals of a regular hexagon 
In this regular convex Hexagon, how many triangles are possible if we consider the intersection points of the diagonals?
I've tried to count the triangles.
First, I counted all the vertices of the hexagon and its diagonals' intersection points(Here 19)
and tried to choose 3 points from the(19C3).
Then, each diagonal has 5 points on them and there are 5 diagonals. So, there should be 5*(5C3) ways that I am overcounting as they don't make any triangle.
My answer is : 19C3 - 5*(5C3). But it is not correct. Why? And what is the correct answer?
 A: Just count the edges of the triangle:

*

*you can't form a triangle inside this hexagon with three sides of the hexagon.

*With two sides you can do it in 6 ways.

*With one side the only nontriangle is if two diagonals perpendicular to the side.  That gives $6\times(3\times 3-1)=48$ ways

*With all diagonals: there are $\frac62(6-3)=9$ diagonals of which there are 3 pairs of parallels, so of the $\binom{9}{3}=84$ ways of selecting three diagonals, we have to exclude $3\times (9-2)=21$ parallel pair of diagonals plus another, and also the $6$ way of selecting all three diagonals from a vertex and $1$ way of three diagonals through the centre.  This leaves $84-21-6-1=56$ ways.

So the total is $6+48+56=110$.
A: This is not complete answer but should be sufficient for you to obtain the answer. The triangles can be divided into three categories:
All 3 vertices on the hexagon
Simply choose $3$ out of the hexagon’s vertices and you will get a triangle with all vertices on the hexagon e.g. $\triangle ABD$.
2 vertices on the hexagon, 1 on a diagonal
Choose $4$ out of the hexagon’s vertices to create a quadrangle. Every quadrangle contains $4$ triangles with $2$ vertices on the hexagon and $1$ on a diagonal e.g. quadrangle $ABCF$ contains $\triangle ABO$, $\triangle BCO$, $\triangle AFO$, and $\triangle CFO$.
1 vertex on the hexagon, 2 on a diagonal
Choose $1$ hexagon’s vertex to be one of the triangle vertices. Then choose $4$ hexagon’s vertices from the remaining $5$ vertices to create a quadrangle. Connect the first vertex to these $4$ vertices and you will see a triangle with one vertex on hexagon’s side and two on a diagonal.
