$p$ points are chosen on each of the three coplanar lines. The maximum number of triangles formed with…
$p$ points are chosen on each of the three coplanar lines. The maximum number of triangles formed with vertices at these points is
$p^3+3 p^2$
$\frac{1}{2}\left(p^3+p\right)$
$\frac{p^2}{2}(5 p-3)$
$p^2(4 p-3)$
Solution
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]
$\begin{aligned}
& =\frac{(3 p) !}{(3 p-3) ! 3 !}-3 \cdot \frac{p !}{(p-3) ! 3 !} \\
& =\frac{3 p(3 p-1)(3 p-2)}{3 \times 2}-\frac{3 \times p(p-1)(p-2)}{3 \times 2} \\
& =\frac{p}{2}\left[9 p^2-9 p+2-\left(p^2-3 p+2\right)\right] \\
& =p^2(4 p-3)
\end{aligned}$