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.
Chat with our AI personalities
There are 5,461,512 such combinations.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
100
44
There are 9C2 = 9*8/(2*1) = 36 2-digit combinations.