There are two answers depending if the numbers can repeat or not
If numbers cannot repeat: For the first digit there are 3 options (3, 2, or 1) for the second digit there are 2 choices remaining since one was already chosen for the first choice and for the last digit there only one choice. Multiplying these choices 3x2x1 you get 6. There are 6 combinations.
You can also solve for this using the permutations formula
n!/(n-r)!
N being the objects you are choosing being 3
R being all the objects in the set being 3
3!/(3-3)!
3!= 3x2x1=6
(3-3)!= 0! (Note 0! is equal to 1)
6/1=6
If numbers can repeat: You have 3 options to choose from for each digit (3, 2, or 1) 3x3x3 or 3^3=27
There are 27 total combinations