Let ABC be a triangle. Consider four points p ₁, p ₂, p ₃, p ₄ on the side AB, five points p₅, p₆, p₇, p₈, p₉ on the side B C , 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 p ₁, p ₂, , p ₁₃ , is _ _ _ _

Step-by-step solution

Points: 4 on AB, 5 on BC, 4 on AC. For a valid pentagon, no 3 vertices can be collinear, so at most 2 points from each side. Only valid distribution (a, b, c) with a+b+c = 5 and each 2 is permutations of (1, 2, 2) : (1,2,2) : 4 1 5 2 4 2 = 4 10 6 = 240 (2,1,2) : 4 2 5 1 4 2 = 6 5 6 = 180 (2,2,1) : 4 2 5 2 4 1 = 6 10 4 = 240 Total = 240 + 180 + 240 = 660 .