QJEEM24Ma6a1c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6a1c Solution: There are 5 points P 1 , P 2 , P 3 , P 4 , P 5 on the side A B…

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QJEEM24Ma6a1c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6a1c Solution: There are 5 points P 1 , P 2 , P 3 , P 4 , P 5 on the side A B…

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 $P_{6}, P_{7}, \dots, P_{11}$ on the side $BC$ and 7 points $P_{12}, P_{13}, \dots, P_{18}$ on the side $CA$ of the triangle. The number of triangles, that can be formed using the points $P_{1}, P_{2}, \dots, P_{18}$ as vertices, is:

Multiple Choice Options

A 751
B 776
C 796
D 771

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Accepted Answer

Number of points on side $A B = 5$

Number of points on side $B C = 6$

Number of points on side $A C = 7$

JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Permutations and Combinations Question 24 English Explanation

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 $P_{1} P_{2} \dots P_{18}$

$\begin{aligned} =^{18} C_{3} -^{5} C_{3} -^{6} C_{3} -^{7} C_{3} \\ = 816 - 10 - 20 - 35 \\ = 751 \end{aligned}$

QJEEM24Ma6a1c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6a1c Solution: There are 5 points P 1 , P 2 , P 3 , P 4 , P 5 on the side A B…

Multiple Choice Options

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