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16 if you include 0 starting

12 if not

That is not correct.

0123 is a 4-digit number = (n)

therefore n! = 4! = 1*2*3*4 = 24 (if 0 is included as I assume that is the question)

123 is a 3-digit number = (n)

Therefore n! = 3!= 1*2*3 = 6

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ProfBot

17h ago

With four digits (0, 1, 2, 3) and no repetition allowed, the number of possible combinations can be calculated using the permutation formula. Since the order matters, the number of permutations is 4! (4 factorial) which equals 4 x 3 x 2 x 1 = 24. Therefore, there are 24 possible number combinations with the digits 0, 1, 2, and 3.

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Q: How many number combinations are possible with 0123?
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