Mathematics · Permutations and Combinations

JEE Main 2025 — 7 April, Evening Shift — Question 28

Let $p$ be the number of all triangles that can be formed by joining the vertices of a regular polygon $P$ of $n$ sides and $q$ be the number of all quadrilaterals that can be formed by joining the vertices of $P$ . If $p+q=126$ , then the eccentricity of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{n}=1$ is :

Option A:
$\frac{\sqrt{7}}{4}$
Option B:
$\frac{1}{\sqrt{2}}$
Correct
Option C:
$\frac{1}{2}$
Option D:
$\frac{3}{4}$

Answer: B

Step-by-step solution

$P={ }^{n} C_{3}=\frac{n(n-1)(n-2)}{6}$

$q={ }^{n} C_{4}=\frac{n(n-1)(n-2)(n-3)}{24}$

$p+q=126$

$\therefore\left(n^{2}-n\right)\left(n^{2}-n-2\right)=3024$

Let $m=n^{2}-n$ $m(m-2)=3024$ $m^{2}-2 m-3024=0$

$m=56$ or $n^{2}-n=56$

$n=8$

$\therefore \quad$ Eqn. of ellipse, $\frac{x^{2}}{6}+\frac{y^{2}}{8}=1$

$e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{1}{\sqrt{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Mathematics
Chapter: Permutations and Combinations
Topic: Problems based on both permutations and combinations