Derangments

 

 

To explain derangements, first you must know what a permutation is.  A permutation of a set of distinct objects is an arrangement of them in a line in some order.  For natural numbers n and r, with r less than or equal to n, an

r-permutation of n objects is a permutation of r of them, that is, an arrangement of r of the objects in a line in some order.  It is important to note that when arranging the r objects into an order, all objects must be used, and no object may be used more than once.  

          A derangement of n distinct objects which have some natural order is a permutation in which no symbol is in its correct position.  We denote a derangement of n objects as Dn .  The number of derangements of n≥1 ordered objects is

                

Dn = n!( 1-(1/1!)+(1/2!)-(1/3!)+…+(-1)n(1/n!) ) .

 

 

For example if we wanted to arrange 3 top 3 finishers in a race on the winners pedestal in such a way that all of them were in the incorrect spot in the medal ceremony, then our objects would be the top 3 finishers of the race, n would equal 3 because there are 3 top finishers to arrange, and our calculation to arrange them all incorrectly would be the derangement:

 

 

D3 = 3!( 1-(1/1!)+(1/2!)-(1/3!) ) = 6(1-1+1/2-1/6) = 6(2/6) = 2 .

 

We can see this if we arrange our finalists in the order of 3,1,2 or 2,3,1.  No other way can we arrange the finalists such that all are in the wrong spot, so our answer of 2 ways is correct.

Let’s test your knowledge about derangements.

 

 

 

 

 

 

Answer the following question about derangements:

 

          The Nebraska Basketball team has made it to the championship game in the Final Four.  However, they must face a team that has studied every game they have played and they know every players weaknesses and strengths.  The Nebraska coach has heard through the grapevine that the opposing team has perfectly matched their starting 5 players to the Nebraska starting 5 in such a way that all of the Nebraska players strengths are evenly matched, so he has decided to rearrange the starting 5 to all different positions to avoid this matchup.  How many ways can he rearrange the starting 5 so that all of them are in a different position than they normally play? (Assume left guard is not equal to right guard, etc.)

 

 

A) 1

B) 34

C) 44

D) 54

E) 120

 

 

 

This page has been created using information from CSE235 taught by Leen-Kiat Soh.  Based on (Rosen 2003), (Goodaire and Parmenter 2002), and (Ross and Wright 1988).

 

 

Jeff Ray

HW 8

Tutorial