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Word Derangements and Arrangements


Date: 11/30/98 at 15:34:52
From: julie
Subject: Derangements

Hi Dr. Math,

I have to write this paper on derangements but I'm a little stuck. 
First you have to find the arrangements on let's say a 3 letter word, 
JOE. JOE has 6 arrangements: J,O,E  J,E,O  O,J,E  O,E,J  E,J,O  E,O,J.
Now to find the derangements, you have to put the letters in 
alphabetical order, which is E J O. This means that E cannot be the 1st 
letter, J cannot be the 2nd letter, and O cannot be the 3rd letter. 
Since there are only two arrangements that meet this criterion, JOE 
has 2 derangements, J,O,E and O,E,J.

I figured out all the derangements for 2, 3, 4, and 5 lettered words 
by listing them, but when I got up to the 6th lettered word, I got 
stuck because a 6 lettered word has 720 arrangements. It'll be 
impossible for me to list all the derangements. Is there a formula 
that I can use to find the derangements of arrangements?


Date: 11/30/98 at 17:00:38
From: Doctor Rob
Subject: Re: derrangements

Yes, there is. The formula is:

   D(n) = n! - n!/1! + n!/2! - n!/3! + n!/4! - ... + (-1)^n*n!/n!

When n = 6, this gives:

   D(6) = 720 - 720 + 720/2 - 720/6 + 720/24 - 720/120 + 720/720
        = 720 - 720 + 360 - 120 + 30 - 6 + 1
        = 265

This formula is gotten by starting with the n! arrangements, then
subtracting those with one object fixed, C(n,1)*(n-1)! = n!/1!, then
adding those with two objects fixed, C(n,2)*(n-2)! = n!/2!, because
they had been subtracted twice. Then we subtract those with three
objects fixed, C(n,3)*(n-3)! = n!/3!, because they had been subtracted
thrice, but added back in thrice. We continue thus until we run out
of objects.

A short way to compute this is to pick the closest integer to n!/e.
Here e is the base of natural logarithms, 2.718281828459045.... If
n = 6, this gives 720/2.718281828459045... = 264.8731976..., so
D(6) = 265, the same answer we got above. The first few answers are
0, 1, 2, 9, 44, 265, 1854, 14833, 133496, and 1334961.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 12/02/98 at 17:40:36
From: julie
Subject: Derangements

Hello Dr. Math,

I wrote to you before to ask if there was a formula to find the 
derangements of arrangements. You replied by sending me this formula:
 
   D(n) = n! - n!/1! + n!/2! - n!/3! + n!/4! - ......+ (-1)^n*n!/n!
 
I understand how to use the formula, but I'm not sure how to derive 
it. You gave me an explanation, but it was a bit confusing. Can you 
please explain it again? Thank you!


Date: 12/03/98 at 10:02:41
From: Doctor Rob
Subject: Re: Derangements

This is an application of the Inclusion-Exclusion Principle. We want 
to know the number of arrangements with exactly zero items in the 
designated spots. That is D(n).

We can count the number C(n,k) of ways to choose k spots out of n to 
be designated; then there are (n-k)! ways to arrange the other n-k 
letters. This makes a total of C(n,k)*(n-k)! = n!/k!. This is, however, 
more than the number of arrangements with k items in their designated 
spots, since some arrangements are counted more than once.

We start with all the arrangements; that is, those with at least zero 
items in the designated spots. There are n!/0! = n! of them. From this 
number we subtract the number we computed above, n!/1!. This is too 
much to subtract, because there are arrangements with two items in 
their designated spots which are counted twice, once for each of the 
two. To compensate, we add back in the then number of arrangements 
with two or more items in the designated spots, namely n!/2!. But we 
have overcompensated, so we subtract the number of arrangements with 
three or more items, and so on.

Example: RUST, alphabetically RSTU. There are 4! = 24 arrangements:

   RSTU, RSUT, RTSU, RTUS, RUST, RUTS, SRTU, SRUT, STRU, STUR, SURT,  
   SUTR, TRSU, TRUS, TSRU, TSUR, TURS, TUSR, URST, URTS, USRT, USTR, 
   UTRS, UTSR.

There are C(4,1) = 4 ways to pick a single letter to be in its 
designated spot, and for each, there are (4-1)! = 3! = 6 ways to 
rearrange the other letters, for a total of 4*6 = 24 = 4!/1!:

   R fixed:  RSTU, RSUT, RTSU, RTUS, RUST, RUTS
   S fixed:  RSTU, RSUT, TSRU, TSUR, USRT, USTR
   T fixed:  RSTU, RUTS, SRTU, SUTR, URTS, USTR
   U fixed:  RSTU, RTSU, SRTU, STRU, TRSU, TSRU

We have to discard these, but some of them appear twice (or more). To
compensate, we add back in the number of arrangements with two letters 
in their designated spots. There are C(4,2) = 6 ways to pick two 
letters to be in their designated spots, and for each there are 2! = 2 
ways to rearrange the other letters, for a total of 6*2 = 12 = 4!/2!:

   R,S fixed:  RSTU, RSUT
   R,T fixed:  RSTU, RUTS
   R,U fixed:  RSTU, RTSU
   S,T fixed:  RSTU, USTR
   S,U fixed:  RSTU, TSRU
   T,U fixed:  RSTU, SRTU

Now we have overcompensated, since some arrangements with two letters
fixed also have three letters fixed, so we have to subtract the number 
of these to compensate. There are C(4,3) = 4 ways to choose the three 
letters to be in their designated spots, and for each there is 1! = 1 
way to rearrange the other letter, for a total of 4*1 = 4 = 4!/3!:

   R,S,T fixed:  RSTU
   R,S,U fixed:  RSTU
   R,T,U fixed:  RSTU
   S,T,U fixed:  RSTU

Again, we have overcompensated, since some (actually all) arrangements
with three letters fixed also have four letters fixed, so we have to 
add back in the number of these to compensate. There is C(4,4) = 1 way 
to choose the four letters to be in their designated spots, and for 
each there is 0! = 1 way to arrange the remaining zero letters, for a 
total of 1*1 = 1 = 4!/4!:

   R,S,T,U fixed:  RSTU

That makes the total:

   D(4) = 4!/0! - 4!/1! + 4!/2! - 4!/3! + 4!/4!
        = 24 - 24 + 12 - 4 + 1
        = 9

The arrangement RSTU with four letters fixed was counted once in the
original tally, subtracted four times, added six times, subtracted four 
times, then added once, for a net count of 0. There are no arrangements 
with exactly three letters fixed. You can verify each arrangement with 
exactly two letters fixed was counted once, subtracted twice, and added 
once, for a net count of 0. You can verify each arrangement with 
exactly one letter fixed was counted once, and subtracted once, for a 
net count of 0. All those with no letter fixed were counted once and 
never subtracted, for a net count of 1. For any arrangement with 
exactly k > 0 letters fixed, the net count is 
C(k,0) - C(k,1) + C(k,2) - ... = (1-1)^k = 0, by the Binomial Theorem.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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