There are three coplanar parallel lines. If any $$p$$ points are taken on each of the lines, the maximum number of triangles with vertices at these points is

Solution :
The number of triangles with vertices on different lines
$$ = {\,^p}{C_1} \times {\,^p}{C_1} \times {\,^p}{C_1} = {p^3}.$$
The number of triangles with 2 vertices on one line and the third vertex on any one of the other two lines
$$ = {\,^3}{C_1}\left\{ {^p{C_2} \times {\,^{2p}}{C_1}} \right\} = 6p \cdot \frac{{p\left( {p - 1} \right)}}{2}$$
∴ the required number of triangles $$ = {p^3} + 3{p^2}\left( {p - 1} \right).$$
Note : The word “maximum” ensures that no selection of points from each of the three lines are collinear.