Question 1183400: Suppose you have n different pairs of socks (n left socks and n right socks, for 2n individual socks total) in your dresser. You take the socks out of the dresser one by one without looking and lay them out in a row on the floor. What is the probability that no two matching socks are next to each other?
Found 2 solutions by robertb, ikleyn:
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! 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
  
and then inserting R1, R2, and R3, one after the other, in ANY of the  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.
------------------------------------------------------------------------
------------------------------------------------------------------------
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.
-----------------------------------------------------------------------
-----------------------------------------------------------------------
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  .
***!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.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
It is DEFINITELY NOT derangement.
Tutor @robertb incorrectly cites the definition of the derangement permutations,
and incorrectly tries to use this notion/conception.
See the referred Wikipedia article on derangement permutations
https://en.wikipedia.org/wiki/Derangement
/////////////
My text below is written in response to notes by @robertb.
-------------------------------------------------------------------
=====================
OK, it looks like I should explain, why I did not solve it and why I think it is not derangement problem.
Let assume that we have two pairs of socks, N1 and N2. Only two pairs, for simplicity.
Let the pair N1 is the pair (1L,1R) : N1 left sock and N1 right sock.
Let the pair N2 is the pair (2L,2R) : N2 left sock and N2 right sock.
We start from this arrangement
1L,1R, 2L,2R. (1)
All your interpretation is based on considering the pairs like
(1L,1R), (2L,2R) (2)
and their derivatives.
To make derangement, I transpose sock 2R from the second pair to the first pair
and transpose sock 1R from the first pair to the second. I get
(1L,2R), (2L,1R). (3)
In your interpretation, two pairs in (3) are deranged: you count it as deranged.
Let's write (3) as a row of socks (without separating them in pairs)
1L, 2R, 2L, 1R.
But now two socks 2R and 2L do belong to one pair and are next to each other, so this arrangement
can not be counted as deranged, and the logical construction is destroyed.
So, your interpretation has an interior contradiction.
It is WHY I could not solve the problem and it is why I think, that the problem,
as it is worded, printed, posted and presented, is DEFECTIVE and can not be solved in terms of derangement.
//////////////
Long time after my previous posts (several months after), I found the solution for closely related TWIN problem in the Internet.
------------------------
Bridget works part-time in a shoe store. Sometimes when it is not busy, she rearranges the shoes for fun.
If she takes six different pairs of shoes and rearranges them in a row, in how many ways can she rearrange them
so that no two shoes match?
------------------------
Find the solution at this link
https://www.quora.com/In-how-many-ways-can-you-rearrange-6-different-pairs-of-shoes-in-a-row-so-that-no-two-shoes-match
The referred text contains hidden parts. When you will read it, open these hidden parts of the text.
ANSWER. The number of ways is 168,422,400 (for n= 6 pairs).
|
|
|
