How many triangles can be formed by the vertices of a regular polygon of $n$ sides? 
How many triangles can be formed by the vertices of a regular polygon of $n$ sides? And how many if no side of the polygon is to be a side of any triangle ?

I have no idea where I should start to think. Can anyone give me some insight ?

Use Combination or Permutation

total no of triangles formed by joining vertices of n-sided polygon
$$= \frac{n(n-1)(n-2)}{6}$$
i.e. selection of 3 points from n points = n(C)3
$\implies$ can also be written as sum of no of triangles formed in the following three cases,
1) no of triangles with only one side common with polygon,
if we take any one side of a n-sided polygon and join its vertices to the remaining vertices, except the vertices adjacent to vertices of the line taken above, we get triangles with only one side as common i.e. for 1 side we get (n-4) triangles $\implies$ n(n-4) triangles for n sides.
case I
2) no of triangles with two sides common,
if we take any one side of a n-sided polygon join its vertex with its opposite vertex required triangle is formed. Hence no of triangles= n
case II
3) triangles with no side common
$$= \text{total - (Case I + Case II)}$$
$$=\left[\frac{n(n-1)(n-2)}{6}\right]-\left[n(n-4) + n\right]$$
$$=\frac{n(n-4)(n-5)}{6}$$
