There are 56 of them and I am not inclined to list them all.
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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
There are only four combinations but there are 8 permutations.
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
Infinitely many. Only 6 with both, the number of 8s and the number of 3s being positive.