JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Permutations and Combinations

For $x$: choose the repeated digit in $9$ ways, choose the remaining $7$ distinct digits from the other $8$ in $\left(\genfrac{}{}{0ex}{}{8}{7}\right)=8$ ways, and arrange the nine digits (one pair) in $\frac{9!}{2!}$ ways, so $x=9\cdot 8\cdot \frac{9!}{2}$. For $y$: choose the two repeated digits in $\left(\genfrac{}{}{0ex}{}{9}{2}\right)=36$ ways, choose the remaining $5$ distinct digits from the other $7$ in $\left(\genfrac{}{}{0ex}{}{7}{5}\right)=21$ ways, and arrange the nine digits (two pairs) in $\frac{9!}{2!2!}$ ways, so $y=36\cdot 21\cdot \frac{9!}{4}$. Hence $\frac{x}{y}=\frac{9\cdot 8\cdot \frac{1}{2}}{36\cdot 21\cdot \frac{1}{4}}=\frac{36}{189}=\frac{4}{21}$, which gives $21x=4y$.