8C3 = 56 of them
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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 125970 combinations and I am not stupid enough to try and list them!
There are 8*7/(2*1) = 28 combinations.
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)
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