Let ABC be a triangle. Consider four points on the side AB, five points on the side BC and four points 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 , is _________.
A pentagon must use 5 points with no three collinear. Since each side of the triangle is a line, at most 2 chosen points can lie on the same side. So we distribute 5 points across three sides with each side contributing at most 2 — the only allowed distributions are in some order.
Three cases by which side contributes only 1 point:
Case 1 — 2 from AB, 2 from BC, 1 from AC: .
Case 2 — 2 from AB, 1 from BC, 2 from AC: .
Case 3 — 1 from AB, 2 from BC, 2 from AC: .
Total: .
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