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

Hint:

Max triangles = Total $C_3$ - (Collinear $C_3$ on each line).

Answer 1

The Correct Option is D

Step 1: Total Points
There are $3$ lines with $p$ points each, giving a total of $3p$ points.
Step 2: Total Combinations
Total ways to choose 3 points = $^{3p}C_{3}$.
Step 3: Subtracting Collinear Points
A triangle cannot be formed if all 3 points are on the same line. There are 3 lines, each having $^pC_{3}$ ways to pick 3 collinear points.
Step 4: Formula Evaluation
$\text{Triangles} = ^{3p}C_{3} - 3 \cdot ^pC_{3} = \frac{3p(3p-1)(3p-2)}{6} - 3 \cdot \frac{p(p-1)(p-2)}{6}$.
Simplifying gives $p^{2}(4p-3)$.
Final Answer: (d)