p points are chosen on each of the three copl | vedclass

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(D) Total number of points is $3p$. To form a triangle,we need to select $3$ points out of $3p$. The total number of ways to select $3$ points is $^{3

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$p^2(4 p-3)$

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(D) Total number of points is $3p$. To form a triangle,we need to select $3$ points out of $3p$. The total number of ways to select $3$ points is $^{3p}C_3$. However,if $3$ points are collinear,they do not form a triangle. There are $3$ lines,each containing $p$ points. For each line,the number of ways to select $3$ collinear points is $^pC_3$. Thus,the number of triangles is $^{3p}C_3 - 3 \cdot ^pC_3$. $= \frac{3p(3p-1)(3p-2)}{6} - 3 \cdot \frac{p(p-1)(p-2)}{6}$ $= \frac{p}{2} [ (3p-1)(3p-2) - (p-1)(p-2) ]$ $= \frac{p}{2} [ (9p^2 - 9p + 2) - (p^2 - 3p + 2) ]$ $= \frac{p}{2} [ 8p^2 - 6p ]$ $= p^2(4p-3)$.

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

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