JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 4)
There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side $$A B$$, excluding $$A$$ and $$B$$, of a triangle $$A B C$$. 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:
751
776
796
771
Explanation
Number of points on side $$A B=5$$
Number of points on side $$B C=6$$
Number of points on side $$A C=7$$
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Number of ways selecting three points from side
$$A B={ }^5 C_3$$
Number of ways selecting three points from side
$$B C={ }^6 C_3$$
Number of ways selecting three points from side
$$A C={ }^7 C_3$$
Total number of triangle possible formed using the points $$P_1 P_2 \ldots P_{18}$$
$$\begin{aligned} & ={ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \\\\ & =816-10-20-35 \\\\ & =751 \end{aligned}$$
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