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\underbar{6.3 Derangements}

Suppose each person in a group of $n$ friends brings a gift to a party.  In how many ways can the $n$ gifts be distributed so that no person receives their own gift.

Let the set of friends = $\{p_1, ..., p_n\}$ where $p_j = $ person $j$.  \hb
Let the set of gifts = $\{g_1, ..., g_n\}$ where $g_j$ = the gift brought by person $j$.

Suppose $f: \{p_1, ..., p_n\}  \rightarrow \{g_1, ..., g_n\}$, \hb
$f(p_k) = g_j$ iff person $p_k$ receives give $g_j$, the gift brought by person $j$. \hb
 If no person receives their own gift. Then $f(p_j) \not= g_j$.

In simpler notation, \hb $f: \{1, ..., n\}  \rightarrow \{1, ..., n\}  $ such that $f(j) \not = j$



Recall: \hb
 a permutation on $ \{1, ..., n\} $ is a bijection $f\{1, ..., n\} \rightarrow \{1, ..., n\} $

Ex:  The permutation 1 2 3 4 5 corresponds to the identity function.

Ex: The permutation 1 3 2 corresponds to the function $f(1) = 1$, $f(2) = 2$, $f(3) = 2$

Defn:  A {\it derangement} of $\{1, ..., n\}$ is a permutation $i_1 i_2 ... i_n$ such that $i_j \not = j$.  I.e, $j$ is not in the $j$th place.


In function notation: $f(j) = i_j$, then if $i_1 i_2 ... i_n$ is a derangement, $f(j) \not= j$


In yet other wording, recall a permuation corresponds to the placement of $n$ non-attacking rooks on an $n \times n$ chessboard.

Ex:  The permutation 1 3 2 corresponds to the following rook placement:

\v

A derangment corresponds to non-attacking rook placement with forbidden positions along the diagonal $(j, j)$, for $j = 1, ..., n$.


Ex:  If rooks are placed on the following $3 \times 3$ chessboard in non-attacking position, then the rook placement corresponds to a derangement if no rook is placed in a spot marked with an $X$.

\v
Thus the derangements of $\{1, 2, 3\}$ are 2 3 1, 2 1 3, 3 1 2, 3 2 1.

\eject

Let $D_n$ = the number of derangements of $\{1, ..., n\}$.

Thm 6.3.1:  For $n \geq 1$, $D_n = n!( 1 - {1 \over 1!} + {1 \over 2!} - {1 \over 3!} + ... + (-1)^n{1 \over n!})$  


Pf:  Use the inclusion and exclusion principle:  If $A_i \subset S$, 

\line{$\overline{\cup A_i} =
|S| - \Sigma_{j = 1}^n |A_j| + \Sigma_{i, j} |A_i \cap A_j| - ... + (-1)^n |A_{1} \cap A_{2} \cap ... \cap A_{n}|.$}
Choose $S$.  What can we count which contains the set of derangements?

Let $S =$ the set of permutations of $\{1, ..., n\}$.  Then $|S| = n!$.

Choose $A_j$ such that the set of derangments = $\overline{\cup A_j}$.
\hb
Let $A_j$ = set of permutations such that $j$ is in the $j$th spot.

Note:  $|A_j| = (n-1)!$ since their is only one choice for the $j$th spot (namely j), leaving $n-1$ terms to permute in the remaining $n-1$ places.


$|A_i \cap A_j| = (n-2)!$

$|A_{i_1} \cap A_{i_2} \cap ... \cap A_{i_k}| = (n-k)!$.


Thus $D_n = n! - \Sigma_{j = 1}^n (n-1)! + \Sigma_{i, j}(n-2)! - ... + (-1)^n (n 
-n)!$

$= \left(\matrix{n \cr 0}\right)  n! -  \left(\matrix{n \cr 1}\right)  (n-1)! +  \left(\matrix{n \cr 2}\right)  (n-2)! - ... +  \left(\matrix{n \cr n}\right)  (-1)^n  0!$

$= n! - {n! \over 1!} + {n! \over 2!} + ... + (-1)^n  {n! \over n!}$  

$= n!(1 - {1 \over 1!} + {1 \over 2!} + ... + (-1)^n {1 \over n!})$  

Recall $\left(\matrix{n \cr k}\right) =$ number of ways to choose $k$ $A_i$'s.


\eject


 

Sidenote:  Finding the number of derangements is often called the hat check problem, because in the old days it was sometimes stated in the following terms:  If $n$ men check their hats, what the probability that the hats are returned so that no one received their own hat.


Recall:  If $E \subset S$, then the probability of $E$ = $P(E) = {|E| \over |S|}$

$S$ = sample space, $E$ = events.

Note:  we assume each outcome is equally likely.


What is the probability that a waiter delivers 4 orders to 4 customers so that no customer receives what they ordered?

Answer:  ${D_4 \over 4!} = 1 - 1 + {1 \over 2} - {1 \over 6} = {1 \over 3}$ 


The probability that a permutation of $\{1, ..., n\}$ is a derangement = ${D_n \over n!} =
1 - {1 \over 1!} + {1 \over 2!} + ... + (-1)^n {1 \over n!}$

Recall from Taylor's expansion from Calculus I, the if $f(x) = $

Recall from Calculus I (Taylor's exp), the $e^{-1} =  \Sigma_{j = 0}^\infty (-1)^j {1 \over !}$.


Thus $e^{-1}$ is a good approximation for the probability of a derangement for $n$ (slightly) large.

Thus the probability that if 

floor, ceiling.

We can derive a recursive formula for $D_n$ (we will look at many recursive formulas in chapter 7).  

Lemma A:  $D_n = (n-1)(D_{n-2} + D_{n-1})$


Note the above formula is a recursive formula as we can determine $D_n$ by calculating $D_k$ for $k < n$.

Note $D_1 = 0$, $D_2 = 1$ (as 2 1 is the only derangement of $\{1, 2\}$).

Thus $D_3 = 2(0 + 1) = 2$, $D_4 = 3(1 + 2) = 9$, $D_5 = 4(2 + 9) = 44$, etc.


Combinatorial proof of lemma A:

$D_n = $ the number of derangements of $\{1, ..., n\}$, a set containing $n$ elements.

$D_{n-1} = $ the number of derangements of $\{1, ..., n-1\}$, a set containing $n-1$ elements.

Let's analyze $D_n$.  The first position of our derangement can be anything except 1.  Thus there are $n-1$ choices for the first position is $n-1$.  Note $n-1$ appears in our formula.  Let's see if we can use this.  

Let $r_2 =$ the number of derangments such that 2 is in the first position.

In general, let $r_k$ = the number of derangments such that $k$ is in the $n$th position.  Note that $r_1 = r_3 = ... = r_{n-1}$ (while $r_n = 0$).  

Then $D_n = r_1 + ... + r_{n-1} = r_{n-1} + ... + r_{n-1} = (n-1)r_{n-1}$.

Thus we have (hopefully) simplified our problem to showing that $r_{n-1} = D_{n-2} + D_{n-1}$


Let $r_{n-1}'$ = the number of derangments such that $n-1$ is in the $n$th position and $n$ is in the $n-1$st position.


Note $r_{n-1}' = D_{n-2}$.

Note $r_n - r_{n-1}' =$ the number of derangments such that $n-1$ is in the $n$th position and $k$ is in the $n-1$st position for some $k \leq n-2$.

We would like to show that $D_{n-1} =$ the number of derangments of $\{1, ..., n-1\}$ such that $n-1$ is in the $n$th position and $k$ is in the $n-1$st position for some $k \leq n-2$.

Let  ${\cal P}_{n}$ the set of derangement where $n-1$ is in the $n$th position and $k$ is in the $n-1$st position for some $k \leq n-2$.  

Let ${\cal D}_{n-1}$ = the set of derangments of $\{1, ..., n-1\}$.  
How can we 

Let's analyze a derangement where $n-1$ is in the $n$th position and $k$ is in the $n-1$st position for some $k \leq n-2$.  

We would like to create a bijection from ${\cal P}_{n}$ to ${\cal D}_{n-1}$

Let $i_1i_2... i_n \in  {\cal P}_{n}$.  Then $i_n = n-1$ and $i_{n-1} = k$ for some $k \leq n-2$.

Create $j_1 j_2 ... j_{n-1}$, a  derangment of $\{1, ..., n-1\}$ by

let $j_l = \cases{ i_l  & if $i_l \not= n$, $1 \leq l \leq n-1$ \cr n-1 & if $i_l \not= n$ } $

Ex:  $25314 \rightarrow 2431$

 
Another (simpler) recurrance relation:

Lemma B:  $D_n = nD_{n-1} + (-1)^n$ for $n \geq 2$


Proof by induction on $n$.

$n = 2$:  $D_2 = 1$ (use definition or Thm 6.3.1)
$2D_1 + (-1)^2 = 2(0) + 1 = 1$.  Thus $D_n = nD_{n-1} + (-1)^n$ holds for $n = 2$.

Suppose $D_{k-1} = (k-1)D_{k-2} + (-1)^{k-1}$ for $k < n$


Then $D_k = (k-1) D_{k-2} + (k-1) D_{k-1} $

Note $D_{k-1} = (k-1)D_{k-2} + (-1)^{k-1}$ implies $(k-1) D_{k-2}  = D_{k-1} - (-1)^{k-1}$ 
 
Thus $D_k = D_{k-1} - (-1)^{k-1} + (k-1) D_{k-1} = k D_{k-1} + (-1) (-1)^{k-1} 
= k D_{k-1} +  (-1)^{k} $ 


\underbar{6.4 Permutations with Forbidden Positions}

Recall in section 6.3, we looked at permutations with forbidden positions 
A derangment corresponds to non-attacking rook placement with forbidden positions along the diagonal $(j, j)$, for $j = 1, ..., n$.

Let $X_j \subset \{1, ..., n\}$ for $j = 1, ..., n$.

Defn:  $P(X_1, X_2, ..., X_n) =$ the set of permutations $i_1i_2...i_n$ of $\{1, ..., n\}$ such that 
$i_j \not\in X_j$.

Defn:  $P(X_1, X_2, ..., X_n) = |P(X_1, X_2, ..., X_n)|$ 

Ex:  $P(X_1, X_2, ..., X_n)$ corresponds to the set of derangements of $\{1, ..., n\}$ if $X_j = \{j\}$.  Thus $D_n = |P( \{1\}, \{2\}, ..., \{n\}|$  

Recall, we can visualize permutations with forbidden positions via $n \times n$ chessboards.

Ex:  Derangements of $\{1, 2, 3\}$


Non-derangement example:  $n = 4$, $X_i = \{j, j+1\}$, $j = 1, 2, 3$, $X_4 = \emptyset$.

$P(X_1, X_2, ..., X_n) = P(\{1, 2\}, \{2, 3\}, \{3, 4\}, \emptyset) = \{3124, 3412, 3421, 4123\}$.

$p(X_1, X_2, ..., X_n) = p(\{1, 2\}, \{2, 3\}, \{3, 4\}, \emptyset) = |\{3124, 3412, 3421, 4123\}| = 4$.

We can use the inclusion-exclusion principle to calculate $p(X_1, X_2, ..., X_n)$ (although in many cases, the computation can be tediously long and beyond computer capabilities for large $n$).

Similar to the proof of Thm 6.3.1,

 $p(X_1, X_2, ..., X_n) =
|S| - \Sigma_{j = 1}^n |A_j| + \Sigma_{i, j} |A_i \cap A_j| - ... + (-1)^n |A_{1} \cap A_{2} \cap ... \cap A_{n}|$

where

Let $S =$ the set of permutations of $\{1, ..., n\}$.  Then $|S| = n!$.


Let $A_j$ = set of permutations $i_1i_2...i_n$ such that $i_j in X_j$ (for a fixed $j$).

Note there are $|X_j|$ ways to place a rook in the $j$th position.  There are $(n-1)!$ ways to place the remaining $n-1$ rooks so that the permutation belongs to $A_j$.  

Thus $|A_j| = |X_j|(n-1)!$.  Thus $\Sigma_{j = 1}^n |A_j| = \Sigma_{j = 1}^n |X_j|(n-1)!= (n-1)!\Sigma_{j = 1}^n |X_j| = r_1(n-1)!$ where $r_1 = \Sigma_{j = 1}^n |X_j|$.

Note $r_1 =$ number of ways to place $1$ nonattacking rooks on an $n \times n$ chessboard so that the rook is in a forbidden position.
  

Let's now look at $A_j \cap A_k$.   $i_1i_2...i_n \in A_j \cap A_k$, then $i_j \in A_j$ and $i_k \in A_k$.  Thus there are $|X_j|$ ways to place a rook in the $j$th position and $|X_k|$ ways to place a rook in the $k$th position.  There are $(n-2)!$ ways to place the remaining $n-1$ rooks so that the permutation belongs to $A_j \cap A_k$.  

Thus $|A_j \cap A_k| = |X_j||X_k|(n-2)!$.  $\Sigma_{i, j} |A_i \cap A_j| = \Sigma_{i, j} |X_j||X_k|(n-2)! =  (n-2)! \Sigma_{i, j} |X_j||X_k|$.  Let $r_2 =  \Sigma_{i, j} |X_j||X_k|$.  
 
Note $r_2 =$ number of ways to place $2$ nonattacking rooks on an $n \times n$ chessboard so that each of the $2$ rooks is in a forbidden position.


Similarly, define $r_k =$ number of ways to place $k$ nonattacking rooks on an $n \times n$ chessboard so that each of the $k$ rooks is in a forbidden position.

Then $\Sigma |A_{i_1} \cap A_{i_2} \cap ... \cap A_{i_k}| = r_k(k-1)!$.

Thm 6.4.1:   $p(X_1, X_2, ..., X_n) = n! - r_1(n-1)! + r_2(n-2)! - ... + (-1)^nr_n$.











 



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