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

Step 1: Understanding the Concept:
This is a permutations and combinations problem. We need to select the digits to be repeated and then arrange them in a 9-digit string.
Step 2: Detailed Explanation:
Finding \(x\):
- Choose 1 digit to be repeated twice: \({}^{9}C_{1}\).
- Choose the remaining 7 distinct digits from the 8 left: \({}^{8}C_{7}\).
- Arrange these \(2 + 7 = 9\) digits: \(\frac{9!}{2!}\).
\[ x = 9 \times 8 \times \frac{9!}{2} = 36 \cdot 9! \] Finding \(y\):
- Choose 2 digits to be repeated twice each: \({}^{9}C_{2}\).
- Choose the remaining 5 distinct digits from the 7 left: \({}^{7}C_{5}\).
- Arrange these \(2 + 2 + 5 = 9\) digits: \(\frac{9!}{2!2!}\).
\[ y = \frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times \frac{9!}{4} = 36 \times 21 \times \frac{9!}{4} = 9 \times 21 \cdot 9! = 189 \cdot 9! \] Comparing \(x\) and \(y\):
\[ \frac{y}{x} = \frac{189 \cdot 9!}{36 \cdot 9!} = \frac{189}{36} = \frac{21 \times 9}{4 \times 9} = \frac{21}{4} \] \[ \Rightarrow 4y = 21x \] Step 3: Final Answer:
The relation is \(21x = 4y\).