JEE Main 2024 — 6 April, Shift 1 — Question 26

${ }^{\mathrm{n}} \mathrm{C}_{1} \mathrm{x}^{\mathrm{n}-1} \mathrm{y}=135$

${ }^{\mathrm{n}} \mathrm{C}_{2} \mathrm{x}^{\mathrm{n}-2} \mathrm{y}^{2}=30$

${ }^{\mathrm{n}} \mathrm{C}_{3} \mathrm{x}^{\mathrm{n}-3} \mathrm{y}^{3}=\frac{10}{3}$

By $\frac{(\mathrm{i})}{(\mathrm{ii})}$

$\frac{{ }^{n} C_{1}}{{ }^{n} C_{2}} \frac{x}{y}=\frac{9}{2}$

By $\frac{\text { (ii) }}{(\text { iii) }}$

$\frac{{ }^{n} C_{2}}{{ }^{n} C_{3}} \frac{x}{y}=9$

By $\frac{(\mathrm{iv})}{(\mathrm{v})}$

$\frac{{ }^{\mathrm{n}} C_{1}{ }^{\mathrm{n}} C_{3}}{{ }^{\mathrm{n}} C_{2}{ }^{\mathrm{n}} C_{2}}=\frac{1}{2}$

$\frac{2 \mathrm{n}^{2}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=\frac{\mathrm{n}(\mathrm{n}-1)}{2} \frac{\mathrm{n}(\mathrm{n}-1)}{2}$

$4 n-8=3 n-3$

$\Rightarrow \mathrm{n}=5$ put in $(\mathrm{v})$

$\frac{x}{y}=9$

$x=9 y$ put in (i) ${ }^{5} \mathrm{C}_{1} \mathrm{x}^{4}\left(\frac{\mathrm{x}}{9}\right)=135$

$x^{5}=27 \times 9$ $\Rightarrow \mathrm{x}=3, \quad \mathrm{y}=\frac{1}{3}$

$6\left(\mathrm{n}^{3}+\mathrm{x}^{2}+\mathrm{y}\right)$

$=6\left(125+9+\frac{1}{3}\right)$ $=806$