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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Permutation and Combination

Let p be the number of all triangles that can be formed by joining...

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 PP. If p+q=126p+q=126, then the eccentricity of the ellipse x216+y2n=1\frac{x^2}{16}+\frac{y^2}{n}=1 is :

Options

  1. A34\frac{3}{4}
  2. B12\frac{1}{2}
  3. C74\frac{\sqrt{7}}{4}
  4. D12\frac{1}{\sqrt{2}}

Correct answer

D. 12\frac{1}{\sqrt{2}}

Step-by-step solution

 Total trangles =⇒=hC3 Total auadrilaterals =hC4=qnC3+nC4=126⇒n+1C4=126⇒n+1=9⇒n=8x216+y2n=1⇒x216+y28=1e=1−816=816=12\begin{aligned} & \text{ Total trangles }=\Rightarrow={ }^{\mathrm{h}} \mathrm{C}_3 \\ & \text{ Total auadrilaterals }={ }^{\mathrm{h}} \mathrm{C}_4=\mathrm{q} \\ & { }^{\mathrm{n}} \mathrm{C}_3+{ }^{\mathrm{n}} \mathrm{C}_4=126 \Rightarrow{ }^{\mathrm{n}+1} \mathrm{C}_4=126 \\ & \Rightarrow \mathrm{n}+1=9 \Rightarrow \mathrm{n}=8 \\ & \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{n}}=1 \Rightarrow \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{8}=1 \\ & \mathrm{e}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}}\end{aligned}
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