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…

Let ABC be a triangle. Consider four points $\displaystyle \mathrm{p}_1, \mathrm{p}_2, \mathrm{p}_3, \mathrm{p}_4$ on the side AB , five points $\displaystyle \mathrm{p}_5, \mathrm{p}_6, \mathrm{p}_7, \mathrm{p}_8, \mathrm{p}_9$ on the side BC , and four points $\displaystyle \mathrm{p}_{10}, \mathrm{p}_{11}, \mathrm{p}_{12}, \mathrm{p}_{13}$ 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 $\displaystyle \mathrm{p}_1, \mathrm{p}_2, \ldots, \mathrm{p}_{13}$, is $\displaystyle \_\_\_\_$