I have determined that if you have N numbers from 0 to N-1 then there are exactly N! (factorial) unique combinations. If this is the case then the unique combinations should be deterministic/formulaic.
For example let's say on a small scale N = 4 that means the unique number of combinations is N! or 4! = 4 * 3 * 2 (excluding the 1 because it's not necessary). If we write this out we can prove it:
0 1 2 3
0 0123 1023 2013 3012
1 0132 1032 2031 3021
2 0213 1203 2103 3102
3 0231 1230 2130 3120
4 0312 1302 2301 3201
5 0321 1320 2310 3210
You can also say that the total number of unique combinations for each starting number 0, 1, 2 & 3 is equal to (N-1)!. So if we look at the first line of unique combinations starting with 0 we can see that there are 6 combinations or 3! = 3 * 2 = 6.
What I can't figure out is how to wrap this up into a formula. For example let K represent the unique combination's position, e.g. K = 13 which would give us 2031 (starting from 0 and counting top-to-bottom and then left-to-right).
So what I am trying to figure out is that given the arguments N (length of combinations), K (the iteration of the combination) and I (index of the number in the combination) how can I get each number of a combination at a given index?
combination(N,K,I)
combination(4,13,2)
24 combinations...
K = 13 or 2031...
3 at index 2 (starting from 0)...
BONUS: I noticed that ALL unique combinations from 0 to N-1 where the number of unique combinations is N! that if you take any two combinations and subtract them from eachother you get a |number| that is ALWAYS divisible by 9--unless N = 1 which is 1.
For example if N = 2, then 2! = 2, which gives us 01 and 10. (10 - 01) = (9 modulo 9) = 0. Or if 4! and we take (3210 - 1302) = (1908 modulo 9) = 0. If you go up to 11! = 39,916,800 you can take (012345678109 - 012345678910) = (|-801| modulo 9) = 0.
What you want is the Factorial number system.
Let's work through your example: $N=4$, $K=13$.
It might be less confusing to explain if we look at combinations of the letters $(A,B,C,D)$ instead. So we have a list of 24 words, starting with $ABCD$ at the $K=0$ position of the list, and $DCBA$ at the $K=23$ position in the list.
As you already noted, there are $N!=24$ combinations of $(A,B,C,D)$ on the list, and there are $(N-1)!=6$ that start with $A$, followed by another 6 that start with $B$, etc.
The one we are looking for, at $K=13$, falls within the third block of 6, because $K = 2\cdot 6+1$. This means that the first letter of our combination is $C$, the third from the list of symbols $(A,B,C,D)$.
The remainder $K' = K-2\cdot6 = 1$ tells us which one of this sub-list of 6 combinations to choose. This sub-list consists of the combinations of the remaining letters $(A,B,D)$. This is exactly the same problem, but with $N'=3$. This list splits into 3 blocks of size $2!$, of which the first block starts with $A$, the second block starts with $B$, the third block starts with $D$.
This time $K' = 0\cdot2+1$, so the combination we are looking for lies in the first block, those which have $A$ as the next Letter. So our combination starts with $CA$, and $K'' = K'-0\cdot2 = 1$ indicates which combination we need to choose from the sub-sub-list of 2.
and so on.
So now to put it together more succinctly into a straightforward procedure:
Represent the number $K$ into the factorial number system, using $N$ 'digits'.
In the example we have $K = 2\cdot3!+0\cdot2!+1\cdot1!+0\cdot0!$, so it is $2010_!$ in factorial representation.
Use the digits in the factorial to select the items from $(A,B,C,D)$.
In the example, we pick $2,0,1,0$:
$(A,B,C,D)$, pick $2$, i.e. $C$.
$(A,B,D)$, pick $0$, i.e. $A$.
$(B,D)$, pick $1$, i.e. $D$.
$(B)$, pick $0$, i.e. $B$.
This gives $CADB$ as the combination at $K=13$.
As for your bonus, Have a look at the Digital Root. You are subtracting two numbers with the same digital root, which means you are subtracting two numbers with the same residue modulo 9, resulting in a number that is 0 modulo 9, i.e. divisible by 9.