Question

Let x be the number of 9 digit numbers formed by taking digits from first 9 natural numbers, where only one digit is repeated twice & y be the number of 9 digit numbers formed from first 9 natural numbers, such that exactly 2 digits repeated twice then

• A. \(x = 27y\)
• B. \(21x = 4y\)
• C. \(5x = 27y\)
• D. \(7x = 27y\)

Hint:

Always perform selection (combinations) first, then arrangement (permutations) to avoid errors in counting problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are forming 9-digit numbers using the set \( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \).
This involves selection of digits and then permutation with repetition.
Step 2: Key Formula or Approach:
For \( x \): 1 digit repeats twice, others are distinct. Total digits needed = 8.
For \( y \): 2 digits repeat twice each, others are distinct. Total digits needed = 7.
Step 3: Detailed Explanation:
Calculation for \( x \):
1. Select 1 digit to repeat: \( \binom{9}{1} \).
2. Select 7 other digits: \( \binom{8}{7} \).
3. Arrange these 9 digits (where 2 are identical): \( \frac{9!}{2!} \).
\[ x = \binom{9}{1} \times \binom{8}{7} \times \frac{9!}{2} = 9 \times 8 \times \frac{9!}{2} = 36 \times 9! \]
Calculation for \( y \):
1. Select 2 digits to repeat: \( \binom{9}{2} \).
2. Select 5 other digits: \( \binom{7}{5} \).
3. Arrange these 9 digits (where two pairs are identical): \( \frac{9!}{2!2!} \).
\[ y = \binom{9}{2} \times \binom{7}{5} \times \frac{9!}{4} = 36 \times 21 \times \frac{9!}{4} = 9 \times 21 \times 9! = 189 \times 9! \]
Comparing \( x \) and \( y \):
\( \frac{y}{x} = \frac{189}{36} = \frac{21}{4} \implies 4y = 21x \).
Step 4: Final Answer:
The relation is \( 21x = 4y \).