If p points are taken on each of the three co | vedclass

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(C) To form a triangle,we need $3$ non-collinear points.Case $1$: One point is chosen from each of the $3$ lines.The number of ways is $^pC_1 	imes ^

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$p^2(4p - 3)$

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(C) To form a triangle,we need $3$ non-collinear points.Case $1$: One point is chosen from each of the $3$ lines.The number of ways is $^pC_1 	imes ^pC_1 	imes ^pC_1 = p^3$.Case $2$: Two points are chosen from one line and one point from one of the other two lines.There are $3$ lines,so we can choose $1$ line to pick $2$ points in $^3C_1$ ways.The number of ways to pick $2$ points from the chosen line is $^pC_2$.The number of ways to pick $1$ point from the remaining $2p$ points is $^{2p}C_1$.Number of triangles = $3 	imes ^pC_2 	imes 2p = 3 	imes \frac{p(p-1)}{2} 	imes 2p = 3p^2(p-1)$.Total number of triangles = $p^3 + 3p^2(p-1) = p^3 + 3p^3 - 3p^2 = 4p^3 - 3p^2 = p^2(4p - 3)$.

If $p$ points are taken on each of the three coplanar parallel lines,what is the maximum number of triangles that can be formed with vertices at these points?

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