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.
To find how many 4-digit numbers equal 19, we need to consider all combinations of four digits (ranging from 0 to 9) that add up to 19. However, since the maximum sum of four digits (9 + 9 + 9 + 9) is 36, the possible combinations are limited. There are no valid 4-digit numbers where the sum of the digits equals 19 while ensuring that the first digit is not zero. Therefore, the answer is that there are no valid 4-digit numbers that equal 19.
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
80
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.
The digit for 19 is simply the number itself, which is composed of two digits: 1 and 9. If you're asking for the individual digits that make up the number 19, they are 1 and 9. In terms of its position in the decimal system, 19 represents the quantity that comes after 18 and before 20.
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)