Question 1183400
UPDATE #2!


To tutor @ikleyn:


You totally didn't understand the strategy I laid out.  The scenario I envisioned was, fixing all left socks on one row, such as in the case of 3 pairs, 
one will have the initial row

{{{L[1]}}}          {{{L[2]}}}              {{{L[3]}}}


and then inserting R1, R2, and R3, one after the other, in ANY of the {{{highlight (FOUR)}}} slots available for insertion, and nowhere else! 
This is for the purpose of producing arrangements such that an R sock will not be either to the left or right of its L match.
 
The arrangement you laid out is not my interpretation, it is YOURS, and yours alone, 
hence you have found contradiction only in your approach, but not mine.  SO you destroyed your own logical construction.


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UPDATE! 

Tutor @ikleyn claims that the problem is NOT about derangement.  But if you looked at and even understood what is presented at 
the wikipedia page she cited (just like I did), then you will realize that IT IS a problem on derangement.  
The strategy is to lay out one from each pair of the 12 pairs of socks on a row and then to insert each of the other 12 socks next to the others laid out.  
We have to count the number of ways that an insertion doesn't produce a "match", i.e., a sock doesn't lie beside its match.  
It then becomes a problem similar to 12 men each of which not getting his correct hat back from the hatstand containing 12 hats. 
Of course, one may argue that the sock might be inserted either to the left or to the right of a fixed sock, but this only doubles the permutations overall. 
Incidentally, this will also double the number of drangements that are possible, so the factor 2 just cancels out.

And to tutor @ikleyn, if you wanted to help a student sincerely, don't just say another tutor is wrong -- give the correct answer and solution! 
Otherwise all your effort is useless and has no value pedagogically.


I think it is safe to say that tutor @ikleyn DOES NOT KNOW the answer, nor has any idea on how to solve the problem.

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This is a problem on the derangements of n = 12 objects, with no sock lying next to its match.

The answer is (!12)/12! = 176,214,841/12! = 0.367879, to 6 d.p. This is almost equal to {{{1/e}}}.

***!n is called the number of drangements of n objects, where none of the n objects are paired correctly with its right match.  
You are referred to https://en.wikipedia.org/wiki/Derangement.