JEE Main 2026 — Permutation Combination Question with Solution

$x$ counts 9-digit numbers using $\{1,2,3,4,5,6,7,8,9\}$ with exactly one digit repeated twice (8 distinct digits, one twice, one absent).
Choose digit appearing twice: 9 ways.
Choose absent digit: 8 ways.
Arrange: $\frac{9!}{2!}$ ways.
$x=9\times 8\times \frac{9!}{2!}=72\times 181440=13043520$.
$y$ counts 9-digit numbers with exactly two digits each repeated twice (7 distinct digits, two appear twice, two absent).
Choose 2 digits for repetition: $\left(\genfrac{}{}{0ex}{}{9}{2}\right)$ ways.
Choose 2 absent digits: $\left(\genfrac{}{}{0ex}{}{7}{2}\right)$ ways.
Arrange: $\frac{9!}{2!\times 2!}$ ways.
$y=36\times 21\times \frac{9!}{4}=756\times \frac{9!}{4}$.
$\frac{x}{y}=\frac{72\times 9!/2}{756\times 9!/4}=\frac{72\times 4}{756\times 2}=\frac{288}{1512}=\frac{4}{21}$.
Therefore $21x=4y$