Count of combinations for case-sensitive string - string

Question

As I remember, the count of combinations equal n!
But, for I example I have string "abc". I want to get all the combinations with different registry: aBc or ABc etc
So, abc is 3 chars. 3! = 1 * 2 * 3 = 6.
But, if I shall try manually do this work - I'll get 8 variations:
1 abc
2 Abc
3 aBc
4 abC
5 ABc
6 aBC
7 AbC
8 ABC
So, it looks like, that answer it 2^3 = 8, but what is 2 ? 3 - is the count of registries in string. what is 2 ? count of registry variants?

Answer 1

If I understand correctly, you want to know for a fixed string how many possible combinations there are with respect to writing the fixed string in mixed capitalization. You are not interested in real permutations of the source string, i.e. you don't wan't to take into account that for abc there is also acb, cab, cba etc. Yes?
If so, for 1 letter we have
a A
for two letters
ab Ab aB AB
and for three letters
abc Abc aBc abC ABc aBC AbC ABC
and so on. If that's the case, then the solution is quite simple if you choose the right underlying model. As you may have noticed, the outcome is regardless of the character sequence we choose - so why not choose all a's:
a A
aa aA Aa AA
aaa aaA aAa aAA Aaa AaA AAa AAA
The pattern is that for each character we have two choices available, either uppercase or lowercase, either set or not set... either 1 or 0 - simply replace a with 0 and A with 1 to get:
0 1
00 01 10 11
000 001 010 011 100 101 110 111
That is actually binary counting! So for n letters the number of possible combinations will be equal to 2^n.

Answer 2

Ah, I think I see what you're saying. If I understand you correctly, you are looking to find all possible ways of capitalizing the letters in a string such that all of the letters aren't of the same case - that is, given abc, you'd produce
abC aBc aBC Abc AbC ABc
But not
abc ABC
Because all letters in those versions have the same case.
If this is what you'd like, the number of ways you can do this in a nonempty string of length n is given by 2n - 2. Intuitively, the rationale behind this is as follows. Given a string of n letters, there are 2n different ways to capitalize all of the letters in that string, since for each character independently of the rest, that letter can be in one of two states (upper case or lower case). If you consider all of those combinations, there are exactly two that you want to disallow - the version where all letters are upper-case, and the version where all letters are lower-case.
In your question, you mentioned that the number of combinations of an n-element sequence is n!. This is not quite right. There are n! permutations of an n-element sequence (assuming each element is distinct). For example, there are 3! = 6 permutations of the sequence abc:
abc acb bac bca cab cba
The fact that there are six ways to capitalize a three-letter sequence without giving all letters the same capitalization and that there are six permutations of abc is a complete coincidence. If you look at more terms of the series, you can see that they only match at two locations (2 and 3):
n = 1 2 3 4 5 6
Permutations (n!) 1 2 6 24 120 720
Mixed-case capitalizations (2^n - 2) 0 2 6 14 30 62
If you allow for more cases than just upper and lower (say, k different versions), then you can generalize this to get the value kn - k, since there are kn different combinations, of which k of them will all have the same capitalization.
Hope this helps!