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 Probability and Statistics Basic Probability and Statistics Math Forum

 January 3rd, 2015, 08:11 AM #1 Newbie   Joined: Jan 2015 From: Turkie Posts: 9 Thanks: 2 Special permutation Hi everyone, I have an urgent problem. I could not find the answer for a long time. Here is the problem: What is the number of permutations of n with the following property: Permutation does not include the number i in the position i. Thank you very much. January 3rd, 2015, 09:57 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra It's not clear to me what $n$ is in this case. January 3rd, 2015, 10:41 AM #3 Newbie   Joined: Jan 2015 From: Turkie Posts: 9 Thanks: 2 n is a posive integer January 3rd, 2015, 10:42 AM   #4
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Hello, talip!

Quote:
 What is the number of permutations of $n$ with the following property: Permutation does not include the number $i$ in the position $i$.

You are dealing with derangements:
$\quad$a permutation of $n$ objects in which
$\quad$none of the objects is in its original position.

Whoever assigned this problem should be aware
$\quad$that the solution is very elusive and intricate.

Here are some values of the derangement function, $d(n)$:

$\qquad \begin{array}{cc} n & d(n) \\ \hline 2 & 1 \\ 3 & 2 \\ 4 & 9 \\ 5 & 44 \\ 6 & 265 \\ 7 & 1854 \\ 8 & 14,\!833 \end{array}$

Good luck! January 4th, 2015, 04:20 AM #5 Newbie   Joined: Jan 2015 From: Turkie Posts: 9 Thanks: 2 Thank you very much for your reply. I need its general function f(n) depending on the independent variable n. I think this problem is more challenging than the one it appears. January 4th, 2015, 10:10 AM   #6
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Hello, talip!

Quote:
 Thank you very much for your reply. I need its general function $f(n)$ in terms of the variable $n.$ I think this problem is more challenging than it appears.

There is a general function for derangements,
but its derivation is quite intricate and convoluted.

I can give you the formula, but it is quite bizarre.

The number of derangements of $n$ objects, $d(n)$, is given by:

$\qquad d(n) \;=\;(Q - 1)^n \quad \text{ where }Q^n \,=\,n!$

Imagine! $\;$ An exponent becomes a factorial!

Example: $\:d(4)$

$d(4) \;=\;(Q-1)^4$

$\qquad =\; Q^4 - 4Q^3 + 6Q^2 - 4Q + 1$

$\qquad =\; 4! - 4(3!) + 6(2!) - 4(1!) + 1$

$\qquad =\;24 - 24 + 12 - 4 + 1$

$\qquad =\;9$

Note that the first two terms always cancel.

Example: $\:d(6)$

$d(6) \;=\;(Q-1)^5$

$\qquad=\;Q^6 - 6Q^5 + 15Q^4 - 20Q^3 + 15Q^2 - 6Q + 1$

$\qquad=\;6! - 6(5!) + 15(4!) - 20(3!) + 15(2!) - 6(1!) + 1$

$\qquad=\;720 - 720 + 360 - 120 + 30 - 6 + 1$

$\qquad=\;265$ January 5th, 2015, 07:37 AM #7 Newbie   Joined: Jan 2015 From: Turkie Posts: 9 Thanks: 2 Thank you very much Soroban. It is what I am looking for. What is the method which you used to find the general form of this function? Again, thank you. May 12th, 2015, 07:35 AM   #8
Newbie

Joined: Jan 2015
From: Turkie

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Quote:
 Originally Posted by talip Thank you very much Soroban. It is what I am looking for. What is the method which you used to find the general form of this function? Again, thank you.
Thank you, the explanatory web page is as follows:
Number of derangements - OeisWiki Tags permutation, special Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Denis Algebra 5 March 13th, 2014 10:54 PM nikkor180 Real Analysis 0 June 17th, 2013 10:56 AM naman Algebra 6 March 6th, 2013 04:29 PM uniquesailor Real Analysis 0 January 27th, 2013 02:35 PM suomik1988 Calculus 2 December 8th, 2010 09:08 PM

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