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How many 4 digit combinations are there for the numbers 0-5 without having the same numbers twice?
Wiki User
∙ 12y ago
There are 360 possible combinations.
Where n = how many numbers are available to choose from and r = the number chosen each time then:
Number of permutations = n! / (n - r)!
= 6! / (6 - 4)!
= (1 * 2 * 3 * 4 * 5 * 6) / (6 - 4)!
= 720 / 2! = 720 / (1 * 2) = 720 / 2 = 360.
(Note: this formula only applies where the order of the numbers is important and repetition is NOT allowed.)
Wiki User
∙ 12y ago
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