Question

There are three coplanar parallel lines; if any \(p\) points are taken on each of the lines, then find the maximum number of triangles with the vertices of these points.

• A. \(p^2(4p - 3)\)
• B. \(p^3 - (4p - 3)\)
• C. \(p(4p - 3)\)
• D. \(p^3\)

Hint:

In combinatorics geometry problems, first find total combinations, then subtract invalid cases.

Answer 1

The Correct Option is A

We must choose vertices from at least two different lines, as all 3 from one line form no triangle. Total points = \(3p\). Total triangles from all points = \(\binom{3p}{3}\). Subtract triangles from same line: \(3 \times \binom{p}{3}\). Simplify: \[ \binom{3p}{3} - 3\binom{p}{3} = p^2(4p - 3) \] Thus, maximum number of triangles = \({p^2(4p - 3)}\).