Infinitely many.
Only 6 with both, the number of 8s and the number of 3s being positive.
8
Since there are 3 numbers to choose from, there will be 2^3 = 8 combinations. This includes the null combinations which you may not wish to include in the count.
8C3 = 56 of them
There are only four combinations but there are 8 permutations.
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
8
Since there are 3 numbers to choose from, there will be 2^3 = 8 combinations. This includes the null combinations which you may not wish to include in the count.
3*5*8 = 120
The answer is 10C3 = 10*9*8/(3*2*1) = 120 combinations.
8C3 = 56 of them
There are only four combinations but there are 8 permutations.
Two . . . . . 38 and 83.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
You can get 50 with 3/8"x3/8" remaining.
6 for 3-digits, 6 for 2-digits, 3 for 1-digits, and 15 for all of the 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
There are 8*7/(2*1) = 28 combinations.