JEE Advanced

Question 39 — JEE Main 28 Jan 2026 (S I)

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  • #11666
    shahkb4
    Keymaster

    Let $\mathrm{S}=\{1,2,3,4,5,6,7,8,9\}$. Let x be the number of 9 -digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let y be the number of 9-digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then, [JEE MAIN 28 Jan 2026 S I]
    (1) $29 x=5 y$
    (2) $45 x=7 y$
    (3) $21 x=4 y$
    (4) $56 x=9 y$

    #11667
    shahkb4
    Keymaster

    Solution

    Using 5,5,1,2,7,8,6,4 digits

    \[ \begin{aligned} & \mathrm{x}=\underset{\substack{\text { Repeating } \\ \text { digit }}}{{ }^{9} \mathrm{C}_{1}} \times \underset{\text { Non repeating }}{{ }^{8} \mathrm{C}_{7}} \times\left(\frac{9!}{2!}\right) \\ & \mathrm{x}=9.8 \frac{9!}{2!}=36(9!) \end{aligned} \]

    Using 1,1,7,7,4,2,5,6,9 digits

    \[ \mathrm{y}=\underset{\substack{\text { repeating } \\ \text { digits }}}{{ }^{9} \mathrm{C}_{2}} \times{ }_{\substack{\text { Non-repeating } \\ \text { digits }}}^{{ }^{7} \mathrm{C}_{5}} \times \frac{9!}{2!2!} \]
    \[ \begin{aligned} & y=\left(\frac{9.8}{2}\right)\left(\frac{7.6}{2}\right)\left(\frac{9!}{2!2!}\right)=9(7) 3.9! \\ & \frac{x}{y}=\frac{36(9!)}{9(7)(3) 9!} \Rightarrow \frac{x}{y}=\frac{4}{21} \Rightarrow 21 x=4 y \text { Answer }(3) \end{aligned} \]
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