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

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) 3p²( p-1) +1

(B) 3p²( p-1)

(C) p²( 4p –3)

(D) none of these

Sol. The number of triangles with vertices on different lines = ^pC₁  × ^pC₁  × ^pC₁ = p³

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

= ³C₁ {^pC₂ × ^2pC₁} $= 6 p .\frac{p(p-1)}{2}$

so, the required number of triangles

= p³ + 3p²(p -1) = p²( 4p – 3)

Hence (C) is the correct answer.