Infinitely many.
Only 6 with both, the number of 8s and the number of 3s being positive.
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8
8C3 = 56 of them
There are only four combinations but there are 8 permutations.
There are 8*7/(2*1) = 28 combinations.
Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56