Question

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

Answer 1

Maximum number of triangles will be formed when no 3 points

on different lines are collinear Total = $3p$ points

Number of ways of selecting 3 points = ${}^{3P}{C}_3$

Number of ways of selecting 3 points

on a single line $=3\times {}^p{C}_3$

$\therefore$ Maximum number oft triangles $-={}^{3p}{C}_3-{3.}^p{C}_3$

$=\frac{\left(3p\right)!}{3!(3p-3)!}-\frac{3.p!}{3!(p-3)!}$

$=\frac{3p(3p-1)(3p-2)(3p-3)!}{3!(3p-3)!}-\frac{3.p(p-1)(p-2)(p-3)!}{3!(p-3)!}$

$=\frac{3p(3p-1)(3p-2)}{6}-\frac{3p(p-1)(p-2)}{6}$

$=\frac{3p(9p^2-6p-3p+2)}{6}-\frac{3p(p^2-3p+2)}{6}$

$=\frac{3p(9p^2-9p+2)-3p(p^2-3p+2)}{6}$

$=\frac{3p(8p^2-6p)}{6}$

$=p(4p^2-3p)=4p^3-3p^2$