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

Answer 1

Answer: (D)

Total number of points in a plane is $3 p$.
$\therefore $ Maximum number of triangles
$={ }^{3 p} C_{3}-3 \cdot{ }^{p} C_{3}$
[Here, we subtract those triangles which points are in a line]
$=\frac{(3 p) !}{(3 p-3) ! 3 !}-3 \cdot \frac{p !}{(p-3) ! 3 !}$
$=\frac{3 p(3 p-1)(3 p-2)}{3 \times 2}-\frac{3 \times p(p-1)(p-2)}{3 \times 2}$
$=\frac{p}{2}\left[9 p^{2}-9 p+2-\left(p^{2}-3 p+2\right)\right]$
$=p^{2}(4 p -3)$