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 on these points is

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

Answer 1

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

Select any three points from total $3p$ points, which can be done ${}^{3p}{C}_3$ ways. But his also includes selection of three colinear points. Now three collinear points from each straight line can be selected in ${}^p{C}_3$ ways. Then the number of triangles is ${}^{3p}{C}_3{-}^{3p}{C}_3=p^2(4p-3)$.