The number of possible 3 digit combinations you can make out of 1-9 with out
repeated digits is:
9C3 = 9!/(3!(9-3)!) = 84
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.
19
9+8+2=19 9+7+3=19 9+6+4=19 There are a limited number of possibilities but there are absolutely loads
It depends which one digit numbers are used. The ten can be any digits from -9 to 9, covering 19 different numbers. If you pick one of those 19 numbers ten times that would be 6131066257801 different possibilities of the number of one digit sets you could make.19*19*19*19*19*19*19*19*19*19=6131066257801
in sigle digit -1 in double digit - 18 in triple digit - 19+(100+19) +19+19 total=195
19 times10,11 (is 2),12,13,14,15,16,17,18,19,21,31,41,51,61,71,81 and 91
71
The first 3 digit number divisible by 19 is 114 (= 19 x 6) The last 3 digit number divisible by 19 is 988 (= 19 x 52) That means that there are 52 - 6 + 1 = 47 three digit numbers that are divisible by 19.
80
42
The first number has 19 possibilities. The second has 18, and the third only 17. 19x18x17=5814 possible lock combinations. (If you could repeat, it would be 19x19x19=6859)
19?