Q44 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
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,
- A$56 x=9 y$
- B$29 x=5 y$
- C$45 x=7 y$
- Correct answerD$21 x=4 y$
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: $\binom{9}{2}$ ways.
Choose 2 absent digits: $\binom{7}{2}$ 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 $21x = 4y$
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: $\binom{9}{2}$ ways.
Choose 2 absent digits: $\binom{7}{2}$ 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 $21x = 4y$
Hence, the correct answer is option: 'D'.