Find the number of triangles whose angular points are at the angular points of a given polygon of n sides, but none of whose sides are the sides of the polygon.

Explanation

A polygon of n sides has n angular points. Number of triangles formed from these n angular points`=.^(n)C_(3)`.
These are comprised of two exclusive cases viz.
(i) atleast one side of the triangle is a side of the polygon.
(ii) no side of the triangle is a side of the polygon.

Let AB be one side of the polygon. if each angular point of the remaining (n-2) points are joined with A and B, we get a triangle with one side AB.
`therefore`Number of triangles of which AB is one side=(n-2)
Likewise, number of triangles of which BC is one side=(n-2) and of which atleast one side is the side of the polygon=n(n-2).
Out of these triangle, some are counted twice. for example, the triangle when C is joined with AB is `DeltaABC`, is taken when AB is taken as one side. again triangle formed when A is joined with BC is counted when BC is taken as one side.
So, the number of triangles of which one side is the side of the triangle.
`=n(n-2)-n=n(n-3)`
`=.^(n)C_(3)-n(n-3)=(1)/(6)n(n-4)(n-5)`