[shahkb4]

shahkb4

3 June, 2026 at 8:22 pm #1105

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, [J
[JEE MAIN 28 Jan 2026 S I]
(1) $29 x=5 y$
(2) $45 x=7 y$
(3) $21 x=4 y$
(4) $56 x=9 y$

Download PDF (Question + Solution)

[shahkb4]

shahkb4

3 June, 2026 at 8:23 pm #1106

Solution

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

\[ \mathrm{x}=\underset{\substack{\text { Repeating } \\ \text { digit }}}{{ }^{9} \mathrm{C}_{1}} \times \underset{\substack{\text { Non repeating } \\ \text { digits }}}{{ }^{8} \mathrm{C}_{7}} \times \underset{\text { arrangement }}{\left(\frac{9!}{2!}\right)} \]

$x=9.8 \frac{9!}{2!}=36(9!)$

Using 1,1,7,7,4,2,5,6,9 digits
$\mathrm{y}=\underset{\substack{\text { repeating } \\ \text { digits }}}{{ }^{9} \mathrm{C}_{2}} \times \underset{\substack{\text { Non-repeating } \\ \text { digits }}}{{ }^{7} \mathrm{C}_{5}} \times \frac{9!}{2!2!}$
$y=\left(\frac{9.8}{2}\right)\left(\frac{7.6}{2}\right)\left(\frac{9!}{2!2!}\right)=9(7) 3.9!$
$\frac{x}{y}=\frac{36(9!)}{9(7)(3) 9!} \Rightarrow \frac{x}{y}=\frac{4}{21} \Rightarrow 21 x=4 y$ Answer(3)

[Author]

Posts