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 :

Mathematics · Permutations and Combinations

JEE Main 2024 — 4 April, Shift 1 — Question 8

There are 5 points $\mathrm{P}_{1}, \mathrm{P}_{2}, \mathrm{P}_{3}, \mathrm{P}_{4}, \mathrm{P}_{5}$ 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 BC and 7 points $\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}$ on the side 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 :

Option A:
776
Option B:
751
Correct
Option C:
796
Option D:
771

Answer: B

Step-by-step solution

There are $18$ points in total:

P_1, P_2, \ldots, P_{18}.

The number of triangles formed by choosing any three points is

\binom{18}{3} = \frac{18 \cdot 17 \cdot 16}{6} = 816.

However, three collinear points do not form a triangle. Points lying on the same side of the triangle are collinear.

Collinear points on each side:

\text{On } AB: 5 \text{ points} \Rightarrow \binom{5}{3} = 10

\text{On } BC: 6 \text{ points} \Rightarrow \binom{6}{3} = 20

\text{On } CA: 7 \text{ points} \Rightarrow \binom{7}{3} = 35

Total degenerate (collinear) triples:

10 + 20 + 35 = 65

Valid triangles:

816 - 65 = 751

\boxed{751}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Permutations and Combinations
Topic: Combinations