Let ABC be a triangle. Consider four points p1, p2, p3, p4 on the side AB, five points p5, p6, p7, p8, p9 on the side BC, and four points p10, p11, p12, p13 on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p1, p2, , p13, is
Answer
660
Solution
A pentagon needs 5 vertices with no three collinear, so at most 2 points can be taken from any one side.
With 5 vertices from 3 sides, the only possible split is 2, 2, 1.
(i) AB: 2, BC: 2, AC: 1 4C2 5C2 4C1 = 6 10 4 = 240
(ii) AB: 2, BC: 1, AC: 2 4C2 5C1 4C2 = 6 5 6 = 180
(iii) AB: 1, BC: 2, AC: 2 4C1 5C2 4C2 = 4 10 6 = 240
Such 5 points lie on the boundary of the triangle with no three collinear, so they always form a (convex) pentagon.
Total = 240 + 180 + 240 = 660
Answer: 660