Let ABC be a triangle. Consider four points on side AB, five points on side BC, and four points on 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 vertices from these points is \_\_\_\_\_\_.
Answer: 660
Solutions · 1
Official solution
A pentagon needs 5 vertices, and since points on the same side of the triangle are collinear, at most 2 vertices can be chosen from any one side (otherwise three or more would be collinear, which cannot be vertices of a simple polygon in the required configuration).
So the only valid distributions of 5 points among the three sides (with counts 4,5,4 available) are of the form (2,2,1) in some order:
\[\binom{4}{2}\binom{5}{2}\binom{4}{1} + \binom{4}{2}\binom{5}{1}\binom{4}{2} + \binom{4}{1}\binom{5}{2}\binom{4}{2}\]
\[= 6\cdot10\cdot4 + 6\cdot5\cdot6 + 4\cdot10\cdot6 = 240 + 180 + 240 = 660\]