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,
- 56x = 9y
- 21x = 4y
- 45x = 7y
- 29x = 5y
Answer
(2) 21x = 4y
Solution
For x: one digit appears twice and 7 other digits appear once (8 different digits, 9 places).
Choose the repeated digit: 9C1 = 9; choose the other 7 digits from the remaining 8: 8C7 = 8
Arrangements = 9!2!
x = 9 8 9!2 = 36 9!
For y: two digits appear twice each and 5 other digits appear once.
Choose the two repeated digits: 9C2 = 36; the other 5 from the remaining 7: 7C5 = 21
Arrangements = 9!2! 2!
y = 36 21 9!4 = 189 9!
xy = 36189 = 421
21x = 4y
Answer: (2)