How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?
Approach Solution - 1
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 =\(900-648=252\)
Approach Solution -2
The total count of integers \(= 999 - 100 + 1 = 900\)
Let \(N\) be the number of integers with no repeated digits.
Since we have \(10\) digits (\(0\) through \(9\)), and the first digit cannot be \(0\), there are \(9\) options for the first digit.
Similarly, for the second digit, there are \(9\) options, and for the third digit, there are \(8\) options.
Thus, the total count of integers with no repeated digits \(= N = 9 × 9 × 8 = 648\)
The count of integers with at least one digit repeated \( = 900 - 648 = 252\)
Therefore, there will be \(252\) integers with at least one repeated digit.
So, the answer is \(252\).
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