There are 5 points P1,P2,P3,P4,P5 on the side AB, excluding A and B, of a triangle ABC. Similarly there are 6 points \[\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}\] on the side \[\mathrm{BC}\] and 7 points \[\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}\] on the side \[\mathrm{CA}\] of the triangle. The number of triangles, that can be formed using the points \[\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}\] as vertices, is:
Solutions · 1
Official solution
Number of points on side AB=5
Number of points on side BC=6
Number of points on side AC=7
Number of ways selecting three points from side
AB=5C3
Number of ways selecting three points from side
BC=6C3
Number of ways selecting three points from side
AC=7C3
Total number of triangle possible formed using the points \[P_1 P_2 \ldots P_{18}\]
\[=18C3−5C3−6C3−7C3=816−10−20−35=751\]