There are 9C3 = 84 combinations.
There are 9C3 = 10*9*8/(3*2*1) = 120 of them.
It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.
45*44*43*42*41/(5*4*3*2*1) = 1,221,759
There are 6C3 = 20 such combinations.
There are 15180 combinations.
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 9C3 = 84 combinations.
There are 1140 5 digit combinations from 1 to 20. 20 combination 3 computes that.
6 of them.
There are 27 possible combinations.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
Allowing repetitions, there are 9 combinations. Without repeated digits, there is only one combination of 3 digits from 3.
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.