Question

Suppose $p$ points are chosen on each of the three coplanar lines. The maximum number of triangles formed with vertices at these points is

• A. $p^3+3p^2$
• B. $\frac{1}{2}(p^3+p)$
• C. $\frac{p^2}{2}(5p-3)$
• D. $p^2(4p-3)$

Answer 1

Answer: (D)

$p^2(4p-3)$

Total number of points in a plane is $3p$.
$\therefore$ Maximum number of triangles $={}^{3p}{C}_3-3\cdot$${}^p{C}_3$
$=\frac{\left(3p\right)!}{(3p-3)!3!}-3\frac{p!}{(p-3)!3!}$
$=\frac{3p(3p-1)(3p-2)}{3\times 2}-\frac{3\times p(p-1)(p-2)}{3\times 2}$
$=\frac{p}{2}[9p^2-9p+2-(p^2-3p+2)]$
$=\frac{p}{2}[8p^2-6p]$

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

Hence. option D is correct.