How many numbers are there in all from 6000 to 6999 having at least one of their digits repeated?

Answer 1

Note that the numbers from 6000 to 6999 comprise all combinations of three digits, with a 6 stuck on the front. We will consider the combinations of these final three digits.

First, let's consider how many three-digit combinations contain one, two or three 7's.

There is one occurrence of three 7's: in the combination '777'.

Now consider two 7's. This can occur in the pattern '77x', '7x7' or 'x77'. For each of these patterns there are 9 possible values for 'x'. Why not 10? Because if x were '7', then it would be '777' which has three, not two 7's. So there are 9 times 3, or 27 ways for there to be exactly two 7's.

Now consider how many occurrences of one 7. The three patterns are '7xy', 'x7y' and 'xy7'. There are 81 combinations of x and y for each of these. (Why not 100? because we are excluding 19 cases where x or y, or both, are 7's.) So there are 81*3, or 243 ways that there can be one 7.

So there are 1 + 27, or 28 ways for there to be at least two 7's. And there are 28 ways for there to be a repeated 8, or a repeated 9, and so on. So there are 9 times 28, or 252 ways for a digit other than 6 to be repeated. Why exclude 6? Because we're going to treat it specially.

All the numbers from 6000 to 6999 begin with a 6. So it is only necessary for a single 6 to occur in the last three digits for it to repeat. And there are 243 + 27 + 1, or 271 ways that one or more 6's can occur, as we saw above.

But wait, we cannot just add 271 to 252, because we would be counting combinations like '6677' or '6767' twice. In fact there are 9 times 3 cases with a repeated 6 and some other repeated digit. (three patterns, and nine possible other digits) So the final count will be 271 + 252 - 27, or 496.

So there are 496 numbers between 6000 and 6999 that have at least one repeated digit.