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Bunuel
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I don't know if this is the correct approach, but:

We have to choose 2, 4 and 5 out of their subset. So 3c3 = 1

We have to choose, out of the remaining 6 numbers, 2 numbers to form a 5 digit integer. So 6c2

We can arrange the chosen numbers in 5! ways

Therefore: 3c3 * 6c2 * 5! = 1800
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Bunuel great question, though believe it should be edited to Five-digit numbers are formed from digits 1 to 9, with no digit repeated rather than Five-digit numbers are formed from digits 1 to 9, with no digit repeated more than once.

No digit repeated more than once implies that a digit can be repeated but at most once eg: 2,2,4,5 & extra digit is allowed or 2,2,4,4,5 but not 2,2,2,4,5.

It may be implied that digits shouldn't be repeated but the phrasing can lead to multiple cases being formed and the answer going beyond 1800.
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