I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
There are 13,244 of them.
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.
Six * * * * * No, that is the number of PERMUTATIONS (not combinations). With 3 numbers, the number of combinations, including the null combination, is 23 = 8. With the three numbers 1,2 and 3, these would be {None of them}, {1), {2), {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
4*3*2*1 = 24 different combinations.
There are: 20C3 = 1140
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 120 permutations and 5 combinations.
3! or Just simply 3x2x1=6 combinations * * * * * Wrong! That is the number of permutations, not combinations. There are 23 combinations: 123, 12, 13, 23, 1, 2, 3 and the null combination.
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210
39*38*37/(3*2*1) = 9139
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