Aspire Faculty ID #18092 · Topic: JEE Main 2026 (28 January Morning Shift) · Just now
JEE Main 2026 (28 January Morning Shift)

Let $ 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,

Solution

$ S = {1, 2, 3, \ldots, 9} $

$ x = {^9C_1} \cdot {^9C_2} \cdot \frac{9!}{2} = \frac{9 \times 8 \times 9!}{2} $

$ y = {^9C_2} \cdot {^7C_2} \cdot \frac{9!}{2! \times 2!} = \frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times \frac{9!}{2! \times 2!} $

$ \Rightarrow \frac{x}{y} = \frac{4}{21} $

$ \Rightarrow 21x = 4y $
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