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  • 3 June, 2026 at 8:19 pm #1091

    Let ABC be a triangle. Consider four points $\mathrm{p}_{1}, \mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{4}$ on the side AB , five points $\mathrm{p}_{5}, \mathrm{p}_{6}, \mathrm{p}_{7}, \mathrm{p}_{8}, \mathrm{p}_{9}$ on the side BC and four points $\mathrm{p}_{10}, \mathrm{p}_{11}, \mathrm{p}_{12}, \mathrm{p}_{13}$ 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 $\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots \mathrm{p}_{13}$, is $\_\_\_\_$ [JEE MAIN 22 Jan 2026 S I]

    3 June, 2026 at 8:19 pm #1092

    Solution


    Required no. of pentagons $=660$ Answer(660)

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