Question

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

Answer 1

Answer: (D)

Total number of points in a plane is 3P.
$ \therefore $ Maximum number of triangles
$ {{=}^{3P}}{{C}_{3}}-{{3.}^{P}}{{C}_{3}} $
(here, we subtract those triangles which points are in a line)
$ =\frac{3P(3P-1)(3P-2)}{3\times 2}-\frac{3\times P(P-1)(P-2)}{3\times 2} $
$ =\frac{P}{2}[9{{P}^{2}}-9P+2-({{P}^{2}}-3P+2)] $
$ ={{P}^{2}}(4P-3) $