Q: 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

Sol. The number of triangles with vertices on different lines = ^{p}C_{1} × ^{p}C_{1} × ^{p}C_{1} = p^{3}

The number of triangles with two vertices on one line and the third vertex on any one of the other two lines

= ^{3}C_{1} {^{p}C_{2} × ^{2p}C_{1}} $= 6 p .\frac{p(p-1)}{2}$

so, the required number of triangles

= p^{3} + 3p^{2}(p -1) = p^{2}( 4p – 3)

Hence (C) is the correct answer.