Question

There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points is

• A. $3p^2(p-1)+1$
• B. $3p^2(p-1)$
• C. $p^2(4p-3)$
• D. none of these

Answer 1

The correct option is C $p^2(4p-3)$

Case 1 : When all points on same line, answer will be zero.

Case 2: One point on each line,

${}_1^{p}C{\times }_1^{p}C{\times }_1^{p}C=p^3$

Case 3: Two points on one line:

${}_1^{3}C$ (line selected) x ${}_2^{p}C$ x ${}_1^{2}C$(one more line selected) x${}_1^{p}C$ = $3p^3-3p^2$

Adding all cases we get $0+p^3+3p^3-3p^2=p^2(4p-3)$