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^2(p-1)` c. `p^2(4p-3)` d. none of these

Options

A. `3n^(2)(n-1)`
B. `3n^(2)(n-1)+1`
C. `n^(2)(4n-3)`
D. none of these

Explanation

The maximum number of triangles = Number of triangles with vertices on different lines + Number of trigangles with 2 vertices on one line and third vertex on any one of the remaining two lines
`=""^(n)C_(1)xx""^(n)C_(1)xx""^(n)C_(1)+""^(3)C_(1)xx(""^(n)C_(2)xx""^(2n)C_(1))=3n^(2)(n-1)`.