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Question 16 : How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?


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Explanatory Answer

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.

  a b c
Possibilities 1 to 9 0 to 9
except a
0 to 9
except a and b
Number of
Possibilities
9 9 8

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


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

Hence, the answer is, "252"

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