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There are 10 numbers 0,1,2,3,4,5,6,7,8,9 in all. Case 1: assuming that the numbers can repeat. The first digit can be chosen in 9 ways (exclude 0). The second in 10 ways(include 0) and so on. Total number of combinations is 9 x 10^7 (that is 9 times 10 to the power 7) ways. Case 2: assuming that the numbers cannot repeat. The first digit in 9 ways, the second digit in 9 ways, the 3rd in 8 ways, the 4th in 7 ways and so on. Total number of permuations is 9 x 9 x 8 x 7 x 6 x 5 x 4 x 3 =1632960 ways

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Q: What are every possible 8 digit combinations of the numbers 0 through 9?
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