As you describe the problem, the condition is you can select any of the 6 numbers with replacement since you have 3 sets of numbers for each value. Per the related link, you have N = 6 and n = 3. The number of combinations is Nn, or 63, or 216. If it is easier to think of the problem in this way, here is another way to look at it. You have 6 choices for the first digit, 6 choices for the second digit, and 6 choices for the third digit. So, you have 6*6*6 = 216 combinations.
To generate all possible three-digit combinations from the given pattern, you can start from 000 and increment by 1 until you reach 999. Here's an example:
000
001
002
...
345
...
998
999
Continue this pattern, incrementing each digit by 1 until all possible combinations have been generated.
It is .012345
it is the menlyn branch
012345
5 is larger than 0 or 1 or 2 or 3 or 4
All of them. That is, 1,000 of them (including zero).
It is .012345
012345 or -543210, if negative numbers are permitted.
012345
Total of 720 combinations:1 --- 12 --- 23 --- 64 --- 245 --- 1206 --- 7207 --- 5040.......Notice that:A --- BC --- DE --- FG --- XIn this sequence, if you wanted to know "X", it would be equal to FxG, and that is the key to this sequence. Why does it happen?Well, if, lets see, for instance, if 2 digits (0 and 1) have a possible of two combinations (10 and 01), then 3 digits (0, 1 and 2) will have a possible of six combinations: 012021And equivelent, but switching 0 with either 1 or 2, making it repeat itself 3 times.Eventually, we will end up doing 2 (number of combinations with 2 digits) times 3 (number of digits of which we want to know the number of unique combinations possible), totallizing 6 (our wanted-to-know number)
543210
012345
it is the menlyn branch
012345
5 is larger than 0 or 1 or 2 or 3 or 4
This is likely an error in your code, post it so we can help
543,210. If you can use them all multiple times, then it is 555,555.
All of them. That is, 1,000 of them (including zero).