<p>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</p>
Step-by-Step Solution
Key Concept: A triangle requires 3 non-collinear points. With 3 parallel lines, you need either 2 points from one line and 1 from another, or 1 point from one line and 2 from the remaining two lines (which must be on different lines to avoid collinearity).
<p><strong>Step 1:</strong> Total ways to choose 3 points from 3p points = C(3p, 3) = 3p(3p-1)(3p-2)/6</p><p><strong>Step 2:</strong> Subtract cases where all 3 points are collinear (on the same line):</p><p>Collinear triples = 3 × C(p, 3) = 3 × p(p-1)(p-2)/6 = p(p-1)(p-2)/2</p><p><strong>Step 3:</strong> Maximum triangles = C(3p, 3) - 3·C(p, 3)</p><p>= 3p(3p-1)(3p-2)/6 - p(p-1)(p-2)/2</p><p>= p[3(3p-1)(3p-2) - 3(p-1)(p-2)]/6</p><p>= p[(9p² - 9p + 2) - (p² - 3p + 2)]/2</p><p>= p[8p² - 6p]/2</p><p>= <strong>p(4p² - 3p)</strong> or <strong>p²(4p - 3)</strong></p><p>∴ Answer: C</p>
Correct Answer: C