JEE Main 2025 Apr 7 Shift 2, Mathematics Q15: Combinations

JEE Main2025Apr 7, Shift 2Mathematics

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 :

A
$\frac{3}{4}$
B
$\frac{1}{2}$
C
$\frac{7}{4}$
D
$\frac{1}{2}$

Solution

Every triangle and every quadrilateral is formed by choosing 3 or 4 vertices out of n, so $p =^{n} C_{3}$ and $q =^{n} C_{4}$ . By Pascal's rule, $^{n} C_{3} +^{n} C_{4} =^{n + 1} C_{4}$ .

Setting $^{n + 1} C_{4} = 126$ : testing $n + 1 = 9$ gives $^{9} C_{4} = 126$ , which checks out, so $n = 8$ .

The ellipse is $\frac{x ^{2}}{16} + \frac{y ^{2}}{8} = 1$ . Since $16 > 8$ , the major axis lies along $x$ with $a^{2} = 16, b^{2} = 8$ .

$e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{8}{16} = \frac{1}{2} = \frac{1}{2}$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q15: Combinations

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