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

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 of the other two lines ${=}^3{C}_1\left\{{}^p{C}_2{\times }^{2p}{C}_1\right\}=6p\frac{p\left(p-1\right)}{2}$

$\therefore$ The required number of triangles$=p^3+3p^2\left(p-1\right)$ .