Question

How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?
[This Question was asked as TITA]

• A. 252
• B. 254
• C. 262
• D. 264

Answer 1

The Correct Option is A

The given set is a set of all three-digit numbers and the number of numbers in the set \(=900\).
The number of three-digit numbers having no digits repeating = \(9×9×8 = 648\)
Required answer =\(900-648=252\)

So, the correct option is (A): \(252\)

Answer 2

The total numbers of integers are = 999 - 100 + 1 = 900
Let N be the number of integers with No Repeated digits
Total 10 digits we have (0, 1 ........, 9)
The 1^st digit can't be 0, as we have 9 digits
For the 2^nd digit, we have 9 digits,
For the 3^rd digit, we have 8 digits
So, Total number of digits with no repetition
⇒ N = 9 × 9 × 8 = 648
Integers with at least one digit repeated,
= 900 - N
= 900 - 648 = 252
Hence, There will be 252 integers with at least one digit repetition.

So, the correct option is (A): \(252\)