Let ABC be a triangle.

Question

Let ABC be a triangle. Consider four points ${p}_{1},{p}_{2},{p}_{3},{p}_{4}$ on the side AB , five points ${p}_{5},{p}_{6},{p}_{7},{p}_{8},{p}_{9}$ on the side $BC$, and four points ${p}_{10},{p}_{11},{p}_{12},{p}_{13}$ 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 ${p}_{1},{p}_{2},\ldots ,{p}_{13}$, is $____$

Answer: 660

Accepted Answer

${}^{4}{C}_{2}\cdot {}^{5}{C}_{2}\cdot {}^{4}{C}_{1}+{}^{4}{C}_{2}\cdot {}^{5}{C}_{2}\cdot {}^{4}{C}_{1}+{}^{4}{C}_{2}\cdot {}^{4}{C}_{2}\cdot {}^{5}{C}_{1}=240+240+180=660$

Solutions · 1