Talk:Derangement

message: d(n) also satisfies the recurrence: d(n) = n*d(n-1) + (-1)^n.
see: http://mathworld.wolfram.com/Derangement.html
I've seen the relation proved by inclusion-exclusion.
DonkeyKong the mathematician (in training) 08:19, 17 July 2006 (UTC)

message: The page about rencontres numbers gives a general formula for the derangements as the closest integer to n! \over e. Is this formula valid for all n? I checked with the first points (ok), and it's obviously valid in the limit. Is there a proof? A corollary of this formula would be a proof that e is irrational. Albmont 13:24, 14 November 2006 (UTC)
replies: text: It is the closest for all natural numbers except n=0. You can always get the correct integer for dn by rounding up at even n and rounding down at odd n. JRSpriggs 05:10, 23 December 2006 (UTC)
replies: text: Probably the formula !n = \left\lfloor\frac{n!}{e} + \frac{1}{2}\right\rfloor should somehow state this, that it doesn't work for n=0?--Thomasda (talk) 15:06, 28 October 2010 (UTC)
replies: text: Yeah, of course, I added it to the article. Paul Breeuwsma (talk) 15:25, 5 December 2010 (UTC)
thread_title:  General formula 

message: Extension:
If all character once : example: ABCDE......
*A008290 Triangle T(n,k) of rencontres numbers (number of *permutations of n elements with k fixed points).[]
:: The proof is an application of the inclusion-exclusion principle.  I'm a bit surprised that this page doesn't say that. Michael Hardy 16:44, 19 June 2007 (UTC)
replies: text:  The details are on the page for Random Permutation Statistics], Maybe we should link there? -Zahlentheorie 09:45, 20 June 2007 (UTC)
thread_title: 1.PART

message: Sorry, not speak english. All corrections thanks! Zerinvary Lajos, Hungary Zlajos OEIS>>zerinvarylajos or zerinvary >>e-mail — Preceding unsigned comment added by 78.92.185.226 (talk) 06:48, 16 April 2009 (UTC)
thread_title: Not speak English

message: :OEIS
#A113899 >>
#A129352 >> 
#A129536 >>   
#Demo>>...mirror of central and multiply of diagonal... (Pascal háromszög tükrözése és szorzás. Minta.)
—Preceding unsigned comment added by Zlajos (talk • contribs) 07:03, 16 April 2009 (UTC)
thread_title: Number parallelogram based on Pascal's triangle (and special mirror of central and multiply of diagonal)

message: Would it be possible to add a section on why this is useful, what are the applications of this math? - Thanks. — Preceding unsigned comment added by 62.60.15.67 (talk) 08:53, 28 August 2012 (UTC)
thread_title: Applications

message: Suppose we have a list A of 2n distinct elements {a1i, a2i; i=1, 2, ..., n}. A dual dearrangement is a permutation of A in which no element a1i or a2i is at any positions 2i and 2i-1, and no two elements a1i and a2i occupy the positions 2j and 2j-1, for any j, in any order.
The method used to derive derangement problem doesn't seem to work for the dual derangement. Any thought?
Mileszhou (talk) 08:12, 20 March 2014 (UTC)
thread_title:  Dual Dearrangement 

message: Does anyone have a proof that the product of all deranged permutations equals the identity?
Darcourse (talk) 15:26, 3 March 2019 (UTC)
replies: text: Doesn't that follow immediately by symmetry? —David Eppstein (talk) 17:05, 3 March 2019 (UTC)

text: I added an answer on my talk. My suggestion that it follows from symmetry assumes that you have a symmetric definition for "the product". In what order are you multiplying them? Also, it's not even true for n=2. —David Eppstein (talk) 21:03, 3 March 2019 (UTC)
replies: text: No matter how you define the product, you will not get the identity when there is an odd number of odd derangements. This appears to happen whenever n = 2 (mod 4); see OEIS:A000387 and note the pattern of odd numbers in it. —David Eppstein (talk) 21:26, 3 March 2019 (UTC)
thread_title:  Derangement product proof