Mathematics · Permutations and Combinations

JEE Main 2026 — 22 January, Morning Shift — Question 24

Let ABC be a triangle. Consider four points $\mathrm{p}_{1}, \mathrm{p}_{2}$ , $p_{3}, p_{4}$ on the side $A B$ , five points $p_{5}, p_{6}, p_{7}, p_{8}, p_{9}$ 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 $\_\_\_\_$

Answer: 660

Numerical answer — enter this value.

Step-by-step solution

Case 1: 2 from $\mathrm{AB}, 2$ from $\mathrm{BC}, 1$

from AC $\binom{4}{2} \cdot\binom{5}{2} \cdot\binom{4}{1}=6 \cdot 10 \cdot 4=240$

Case 2: 2 from $\mathrm{AB}, 1$ from $\mathrm{BC}, 2$

from AC $\binom{4}{2} \cdot\binom{5}{1} \cdot\binom{4}{2}=6 \cdot 5 \cdot 6=180$

Case 3 : 1 from $\mathrm{AB}, 2$ from $\mathrm{BC}, 2$

from AC $\binom{4}{1} \cdot\binom{5}{2} \cdot\binom{4}{2}=4 \cdot 10 \cdot 6=240$

$240+240+180=660$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Permutations and Combinations
Topic: Problems based on both permutations and combinations