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
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?
To calculate the 2 digit numbers divisible by 19, you simply start out by multiplying 19 by 2, then by 3, etc., until you reach a number that is 99 or less. The reason you have do it until you reach a number that is 99 is because after 99 you have a 3 digit number. 19 x 2 = 36 19 x 3 = 57 19 x 4 = 76 19 x 5 = 95 So, the 2 digit numbers divisible by 19 are 36, 57, 76, 95