Question

How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?​

Answer 1

Answer:

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 first digit can't be 0, we have 9 digits

For the second digit, we have 9 digits,

For the third digit, we have 8 digits

Total number of digits with no repetition

⇒ N = 9 × 9 × 8 = 648

Integers with at least one digit repeated

⇒ 900 - N

⇒ 900 - 648 = 252

∴ There will be 252 integers with at least one digit repetition.

Answer 2

The total number of integers from 100 to 999 is 999 - 99 = 900.

Of these 900 integers, if we could find the number of integers which do not have repetitions in them, we can also find the number of integers that have repetitions.

Let's find the number of integers that between 100 and 999, which do not have repetitions.

The number of integers that between 100 and 999, which do not have repetitions = 9 × 9 × 8

Therefore, the number of integers that have repetitions = 900 - 9 × 9 × 8

= 9 (100 - 9 × 8)

= 9 (28)

= 180 + 72 = 200 + 52 = 252

Answer

252