I already know what is the number of Enigma combinations and how did they got to this number :
$${5!\over(5-3)!}\cdot26^3\cdot{26!\over(26-20)!\cdot2^{10}\cdot10!}=158,962,555,217,826,360,000$$
But let me please tell you how I see the right side part of the equation : (the $26!...$ part)
We should choose 10 pairs of letters from $a-z$.
Ok so we have these 10 pairs in braces:
$$(26*25)*(24*23)*(22*21)*(20*19)*(18*17)*(16*15)*(14*13)*(12*11)*(10*9)*(8*7) = 560127029342507827200000$$`
let's divide with the order of choosing those braces $$(..)*(..)*....(..)$$ which is $10!$
And then divide again with internal combination within each brace $(..)$ becuase $AB$ is like $BA$ (10 times) so it's $2^{10} = 1024$
so it's
$$ \frac{560127029342507827200000}{(10! * 2^{10})} = 150738274937250$$
OK.
But The thing which helped me understand this is the braces in $(26*25)*(24*23)...$
Question
But what if I remove(!) the parenthesis ? sure the result will remain the same but here - my understanding is changing.
Let's see :
$26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7$ = combinations of arranging 20 letters = $560127029342507827200000$
let's divide by the order of choosing which is 20! and we get : 230230
ERR.... let's stop here - it's already an illogical too small value to continue.
What am I missing ( logically)?
How do I treat this problem without braces ?
Again - braces of pairs helped me understand that I should divide with $paris!$ and then divide again with internal order with each pair
But If i remove the braces - I get (in terms of words) a new story.
So now I'm having a new story of selecting 10 letters from 26 and now im dividing it by 10! (since I dont care about order) but where does 2^10 comes into play ?
This last number is the number of Steckerbrett options. Here we choose 10 pairs out of the 26 letters to connect. This connection is symmetric so AC and CA are the same connection, e.g. Also, the order of the pairs is irrelevant, so for two pairs out of 6 letters the pairings $(AB)(CD)EF$ and $(CD)(AB)EF$ are the same.
So to generate all pairings we just take any order of the 26 letters, from left to right, and use the pairings given by letters 1 and 2, letters 3 and 4, up to letters 19 and 20, 10 pairs in all. The last $6 = 26- 6$ letters are unpaired.
It's clear we get all possible pairings in this way, but we generate equivalent pairings when we interchange the two letters in any pair, and there are 10 pairs so $2^{10}$ (because we can change the order in any one of them independently) many orders give rise to the same pairing. Also the order of the pairs does not matter for the setting either, so we can interchange them, so we get $10!$ shuffles of the $10$ pairs that are also equivalent. Finally, the order of the 6 unpaired letters at the end is also irrelevant. This makes for a factor $6!$ of overcounting. There are no other ways that an sequence of $26$ letters gives rise to the same settings as another such sequence: they must differ in order of pairs, order within a pair, or order of the unpaired letters.
So every sequence has $2^{10} \cdot 6! \cdot 10!$ many equivalents.
So the total number of classes (so different settings), is
$$\frac{26!}{2^{10} \cdot 6! \cdot 10!}$$
Note that the $\frac{26!}{6!}$ is just your term
$$(26 \cdot 25) \cdot (24 \cdot 23) \ldots \cdot (8 \cdot 7)$$
which is indeed the number of ordered sequences of $20$ letters from a $26$ letter alphabet. You can directly divide that number with the same factors $2^{10} \cdot 10!$ to account for the facts that the order within a pair can be interchanged (independently) and all $10$ pairs can be interchanged in $10!$ ways. We shouldn't divide that number by $20!$ because the undoes all of the ordering, and we get just the number of $20$ letter sets from the alphabet, which indeed equals
$$\binom{26}{20} = \frac{26!}{20! \cdot 6!}$$
But we need the order to make the pairs and compensate for their unorderedness later, as we did.