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The first digit can be any one of three ... 6, 7, or 8.

The second digit can be any one of the three remaining.

The third digit can be either one of the two remaining.

The fourth digit can be either the remaining digit, or else not there ... two choices.

The total number of possibilities is (3 x 3 x 2 x 2) = 36 of them.

That doesn't seem like enough, but that's my analysis and I'm sticking to it.

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Q: How many numbers greater than 600 can be formed with the digits 5 6 7 and 8 if no digit is used more than once in a number?
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