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If numbers can be repeated and zeroes are allowed to lead, this is simply all natural numbers in the set {0000 - 9999}, for a total of 10,000 possible combinations. If numbers cannot be repeated, this becomes a permutation problem; out of 10 possible digits, permute four of them. This evaluates as:

nPr = n! / (n-r)!

10P4 = 10! / (10-4)!

10P4 = 1 * 2 * 3 * ... * 6 * 7 * 8 * 9 * 10 / 1 * 2 * 3 * 4 * 5 * 6

10P4 = 7 * 8 * 9 * 10

10P4 = 5040

There are thus 5,040 possible combinations of four of the digits in 0-9 if any digit is not used twice and 0 is allowed to be a leading digit.

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Q: Possible four digit numbers using 0-9?
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