JEE Main 2026 — Mathematics Algebra

$x$x counts 9-digit numbers using $\{1,2,3,4,5,6,7,8,9\}${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!}$2!9!​ ways.

$x = 9 \times 8 \times \frac{9!}{2!} = 72 \times 181440 = 13043520$x=9×8×2!9!​=72×181440=13043520.

$y$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: $\binom{9}{2}$(29​) ways.

Choose 2 absent digits: $\binom{7}{2}$(27​) ways.

Arrange: $\frac{9!}{2! \times 2!}$2!×2!9!​ ways.

$y = 36 \times 21 \times \frac{9!}{4} = 756 \times \frac{9!}{4}$y=36×21×49!​=756×49!​.

$\frac{x}{y} = \frac{72 \times 9!/2}{756 \times 9!/4} = \frac{72 \times 4}{756 \times 2} = \frac{288}{1512} = \frac{4}{21}$yx​=756×9!/472×9!/2​=756×272×4​=1512288​=214​.

Therefore $21x = 4y$21x=4y