Question

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:

• A. \(21x=4y\)
• B. \(45x=7y\)
• C. \(56x=9y\)
• D. \(29x=5y\)

Hint:

In permutation problems with repetition, always divide by factorials of repeated elements to avoid overcounting.

Answer 1

The Correct Option is A

Concept: This is a permutations problem involving repetition of digits. Carefully count the number of distinct permutations under the given repetition constraints.
Step 1: Find \( x \) Choose the digit to be repeated: \[ \binom{9}{1} \] Choose the remaining 7 distinct digits from the remaining 8 digits: \[ \binom{8}{7} \] Total permutations of 9 digits with one digit repeated twice: \[ x=\binom{9}{1}\binom{8}{7}\frac{9!}{2!} \]
Step 2: Find \( y \) Choose the two digits to be repeated: \[ \binom{9}{2} \] Choose the remaining 5 distinct digits from the remaining 7 digits: \[ \binom{7}{5} \] Total permutations of 9 digits with two digits repeated twice each: \[ y=\binom{9}{2}\binom{7}{5}\frac{9!}{2!2!} \]
Step 3: Compare \( x \) and \( y \) \[ \frac{x}{y} =\frac{\binom{9}{1}\binom{8}{7}\cdot 2!}{\binom{9}{2}\binom{7}{5}} =\frac{4}{21} \] \[ \Rightarrow 21x=4y \]