There are 5 points ${P}_{1},{P}_{2},{P}_{3},{P}_{4},{P}_{5}$ on the side $AB$, excluding $A$ and $B$, of a triangle $ABC$. Similarly there are 6 points ${P}_{6},{P}_{7},\ldots ,{P}_{11}$ on the side BC and 7 points ${P}_{12},{P}_{13},\ldots ,{P}_{18}$ on the side CA of the triangle. The number of triangles, that can be formed using the points ${P}_{1},{P}_{2},\ldots ,{P}_{18}$ as vertices, is : [JEE Main 2024, 4 Apr (Shift 1)]

Question

YESPYQ · Accepted Answer

Correct answer: b) 751. Total ways of selecting $3$ points out of $18$ is ${ }^{18} \mathrm{C}_3$. Total ways of selecting $3$ points from $\mathrm{P}_1, \mathrm{P}_2, \mathrm{P}_3, \mathrm{P}_4, \mathrm{P}_5$ is ${ }^5 \mathrm{C}_3$. Total ways of selecting $3$ points from $\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}$ is ${ }^6 \mathrm{C}_3$. Total ways of selecting $3$ points from $\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}$ is ${ }^7 \mathrm{C}_3$. Hence,The number of triangles, that can be formed using the points $\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}$ is ${ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 $ $=751$

There are 5 points \({P}_{1},{P}_{2},{P}_{3},{P}_{4},{P}_{5}\) on the side \(AB\), excluding \(A\) and \(B\), of a trian…

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