\input epsf
\input graphicx
\magnification 1850
\parindent 0pt
\parskip 10pt
\pageno=1
\hsize 7.2truein
\hoffset -0.35truein
\voffset -0.3truein
\vsize 10truein
\def\u{\vskip -10pt}
\def\v{\vfill}
\def\s{\vskip -5pt}
\def\r{\vskip -4pt}
\def\t{\vskip 10pt}
\def\Z{\cal Z}
\def\hb{\hfil \break}
\def\hr{\vskip 5pt \hrule }
\def\S5{\Sigma_{n=1}^5}

\underbar{6.5 Another  Forbidden Position Problem}

{\bf Goal:}  To {\bf derive} a  formula for counting the number of permutations with relative forbidden positions.

Ex:  Suppose children 1, 2, 3, 4, and 5 sit in a row in class.  Children 1 and 2 cannot sit next to each other or they will cause trouble.  

The order in which the children sit corresponds to a permutation of $\{1, 2, 3, 4, 5\}$.  If 1 is in the $i$th spot, then 2 cannot be in the $i-1$st spot or the $i+1$th spot.  Thus the pattern 21 or 12 cannot appear in our permutation.  This is called a relative forbidden position as certain positions for the placement of 2 are forbidden, but these forbidden positions depend on the placement of 1.

We will focus on the relative forbidden position problem in which 

Let $Q_n = $ the number of permutations of $\{1, 2, ..., n\}$ in which none of the patterns 12, 23, 34, ..., $(n-1)n$ occurs.

Thm 6.5.1 
$Q_n= n!  -  \left(\matrix{n-1 \cr 1}\right)  (n-1)! +  \left(\matrix{n -1 \cr 2}\right)  (n-2)! - ... +  \left(\matrix{n-1 \cr n-1}\right)  (-1)^{n-1}  1!$
   

Proof:  Use inclusion-exclusion principle.

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


Let $A_j$ = set of permutations which contain the pattern $j(j+1)$ for $j = 1, ..., {\bf n-1}$.

To determine $|A_j|$, we can first look at all the permutations of $\{1, 2, ..., j, j+ 2, ..., n\}$.  Since this set has $n-1$ elements, the number of   permutations of $\{1, 2, ..., j, j+ 2, ..., n\} = (n-1)!$.  Note that these permutations are in 1-1 correspondence with $A_j$ as any permutation in $A_j$ can be formed from a  permutation of $\{1, 2, ..., j, j+ 2, ..., n\}$ by inserting $j+1$ right after $j$.  Thus  $|A_j| = (n-1)!$ 


Claim: $|A_i \cap A_j| = (n-2)!$ if $1 \leq i < j \leq n$

Suppose $|i - j| \geq 2$.  In this case, $A_i \cap A_j$ is in 1-1 correspondence with permutations of a set of $n-2$ elements as we can create any permutation of $\{1, 2, ..., i, i+2, ..., j, j+ 2, ..., n\}$ by taking a permutation in $A_i \cap A_j$ and deleting $i+1$ and $j+1$.  Thus  $|A_j| = (n-2)!$ in this case. 

Suppose $|i - j| = 1$.  Thus $j = i+1$ (since we assumed $i < j$).   $A_i \cap A_{i+1}$ is the set of  permutations which contain the pattern $i~ i + 1$ and the pattern $i+1~ i+2$.  Thus  $A_i \cap A_{i+1}$ is the set of  permutations which contain the pattern $i ~i + 1~ i+2$. Thus $A_i \cap A_{i+1}$ is in 1-1 correspondence with permutations of a set of $n-2$ elements as we can create any permutation of $\{1, 2, ..., i, i+3, ..., n\}$ by taking a permutation in $A_i \cap A_{i+1}$ and deleting $i+1$ and $i+2$.  Thus  $|A_i \cap A_{i+1}| = (n-2)!$ in this case as well. 



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


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

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

\end
Recall $\left(\matrix{n-1 \cr k}\right) =$ number of ways to choose $k$ $A_i$'s where $1 \leq i \leq n-1$.










 



\end