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\begin{document}
\section*{Math 222 Fall 2022, Lecture 13: Introduction}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/22fco}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/22fco}}}

\section{Introduction (cont'd)}

\setcounter{subsection}{6}

\subsection{Counting permutations: an introduction}

\setcounter{subsubsection}{3}

\setcounter{theo}{7}

\subsubsection{The one-line notation}

If $\sigma$ is a permutation of the set $\left[  n\right]  $ (for some
$n\in\mathbb{N}$), then we can write $\sigma$ in two-line notation as follows:%
\[
\sigma=\left(
\begin{array}
[c]{cccc}%
1 & 2 & \cdots & n\\
\sigma\left(  1\right)  & \sigma\left(  2\right)  & \cdots & \sigma\left(
n\right)
\end{array}
\right)  .
\]
But the top row here carries no useful information: We already know that the
elements of $\left[  n\right]  $ are $1,2,\ldots,n$, and it is not exactly
unreasonable to list them in increasing order. Thus, we can omit it, and write
only the bottom row, i.e., the $n$-tuple%
\[
\left(  \sigma\left(  1\right)  ,\ \sigma\left(  2\right)  ,\ \ldots
,\ \sigma\left(  n\right)  \right)  \in\left[  n\right]  ^{n}.
\]
This is the so-called \textbf{one-line notation} of $\sigma$:

\begin{definition}
Let $n\in\mathbb{N}$. Let $\sigma$ be a permutation of $\left[  n\right]  $.
Then, the \textbf{one-line notation} of $\sigma$ means the $n$-tuple $\left(
\sigma\left(  1\right)  ,\ \sigma\left(  2\right)  ,\ \ldots,\ \sigma\left(
n\right)  \right)  $.
\end{definition}

For instance, the six permutations of $\left[  3\right]  $ have one-line
notations%
\[
\left(  1,2,3\right)  ,\ \ \ \ \left(  1,3,2\right)  ,\ \ \ \ \left(
2,1,3\right)  ,\ \ \ \ \left(  2,3,1\right)  ,\ \ \ \ \left(  3,1,2\right)
,\ \ \ \ \left(  3,2,1\right)  .
\]
These are precisely all possible triples (i.e., $3$-tuples) of distinct
elements of $\left[  3\right]  $ that contain each element of $\left[
3\right]  $. This pattern persists for \textquotedblleft higher values of
$3$\textquotedblright:

\begin{proposition}
Let $n\in\mathbb{N}$. Then:

\textbf{(a)} If $\sigma$ is any permutation of $\left[  n\right]  $, then the
one-line notation of $\sigma$ is an $n$-tuple of distinct elements of $\left[
n\right]  $ that contains each element of $\left[  n\right]  $.

\textbf{(b)} If $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is an $n$-tuple of
distinct elements of $\left[  n\right]  $ that contains each element of
$\left[  n\right]  $, then there exists a unique permutation $\sigma$ of
$\left[  n\right]  $ such that $\left(  i_{1},i_{2},\ldots,i_{n}\right)  $ is
the one-line notation of $\sigma$.

\textbf{(c)} Let%
\[
T_{n}=\left\{  n\text{-tuples of distinct elements of }\left[  n\right]
\text{ that contain each element of }\left[  n\right]  \right\}  .
\]
Then, the map%
\begin{align*}
\left\{  \text{permutations of }\left[  n\right]  \right\}   &  \rightarrow
T_{n},\\
\sigma &  \mapsto\left(  \text{one-line notation of }\sigma\right)  =\left(
\sigma\left(  1\right)  ,\ \sigma\left(  2\right)  ,\ \ldots,\ \sigma\left(
n\right)  \right)
\end{align*}
is well-defined and a bijection.
\end{proposition}

\begin{proof}
This is an easy exercise in understanding the definitions (and basic set
theory). Most importantly, a map $\sigma:\left[  n\right]  \rightarrow\left[
n\right]  $ is uniquely determined by its list of values $\left(
\sigma\left(  1\right)  ,\ \sigma\left(  2\right)  ,\ \ldots,\ \sigma\left(
n\right)  \right)  $, and furthermore this map $\sigma$ is

\begin{itemize}
\item injective if and only if the list $\left(  \sigma\left(  1\right)
,\ \sigma\left(  2\right)  ,\ \ldots,\ \sigma\left(  n\right)  \right)  $ is a
list of \textbf{distinct} elements;

\item surjective if and only if the list $\left(  \sigma\left(  1\right)
,\ \sigma\left(  2\right)  ,\ \ldots,\ \sigma\left(  n\right)  \right)  $
contains each element of $\left[  n\right]  $.
\end{itemize}

See the 2019 notes (Proposition 1.7.12) for details.
\end{proof}

The main use of the one-line notation (besides being short) is that it allows
us to manipulate permutations as tuples. We will eventually see some instances
of that (e.g., in Lecture 28, proof of Proposition 4.4.5).

\subsubsection{Short-legged permutations}

Now let us count a few more classes of permutations.

\begin{definition}
Let $n\in\mathbb{N}$. A permutation $\sigma$ of $\left[  n\right]  $ will be
called \textbf{short-legged} if each $i\in\left[  n\right]  $ satisfies
$\left\vert \sigma\left(  i\right)  -i\right\vert \leq1$.
\end{definition}

How many such permutations does $\left[  n\right]  $ have?

For example, the short-legged permutations of $\left[  3\right]  $ are (in
one-line notation)%
\[
\left(  1,2,3\right)  ,\ \ \ \ \left(  1,3,2\right)  ,\ \ \ \ \left(
2,1,3\right)  .
\]
So there are $3$ of them. Based on experiments, we suspect:

\begin{proposition}
Let $n\in\mathbb{N}$. Then,
\[
\left(  \text{\# of short-legged permutations of }\left[  n\right]  \right)
=f_{n+1}.
\]
(Again, the Fibonacci numbers.)
\end{proposition}

\begin{proof}
See the 2019 notes (\S 1.7.5) or (more fun!) figure it out yourself (what is
the relation between short-legged permutations of $\left[  n\right]  $ and
other combinatorial objects that are counted by Fibonacci numbers?).
\end{proof}

\begin{exercise}
How many short-legged derangements does $\left[  n\right]  $ have?
\end{exercise}

\subsubsection{Long-legged permutations}

\begin{definition}
Let $n\in\mathbb{N}$. A permutation $\sigma$ of $\left[  n\right]  $ will be
called \textbf{long-legged} if each $i\in\left[  n\right]  $ satisfies
$\left\vert \sigma\left(  i\right)  -i\right\vert >1$.
\end{definition}

Again, we can ask: How many are there? Here is a table:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|c|c|c|}\hline
$n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$\\\hline
$\ell_{n}$ & $1$ & $0$ & $0$ & $0$ & $1$ & $4$ & $29$ & $206$ & $1708$ &
$15702$\\\hline
\end{tabular}
\ \ ,
\]
where $\ell_{n}$ denotes the \# of long-legged permutations of $\left[
n\right]  $.

On the OEIS, the sequence $\left(  \ell_{0},\ell_{1},\ell_{2},\ldots\right)  $
is \href{https://oeis.org/A001883}{Sequence A001883}, and you can find two
recurrent equations for it, one of which is%
\begin{align*}
\ell_{n}  &  =n\ell_{n-1}+4\ell_{n-2}-3\left(  n-3\right)  \ell_{n-3}+\left(
n-4\right)  \ell_{n-4}\\
&  \ \ \ \ \ \ \ \ \ \ +2\left(  n-5\right)  \ell_{n-5}-\left(  n-7\right)
\ell_{n-6}-\ell_{n-7}\ \ \ \ \ \ \ \ \ \ \text{for all }n\geq7
\end{align*}
(not the nicest formula, but beats checking all the $n!$ permutations of
$\left[  n\right]  $ for sure!). This seems to be the most direct expression
we have for $\ell_{n}$. No closed-form formula is known (or expected to
exist). Not every counting problem has a good answer!

\section{Binomial coefficients}

Let us continue our exploration of BCs (binomial coefficients) and get a
little bit more systematic about proving identities. In particular, we shall
(later in this chapter) prove the Chu--Vandermonde identity\footnote{Note: you
are already free to use it on homeworks and midterms.}, and derive various
consequences from it. We will also learn some methods for proving identities.

\subsection{Revisiting the alternating sum of a row in Pascal's triangle}

Recall the following fact (Proposition 1.3.23 in Lecture 7):

\begin{proposition}
\label{prop.binom.altsum=0}We have%
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}=\left[  n=0\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}.
\]

\end{proposition}

We already proved this (by applying the binomial formula to $1$ and $-1$). We
shall now give two more proofs of this proposition. Each proof will exemplify
an important technique.

\subsubsection{Telescoping sums}

The first new proof of Proposition \ref{prop.binom.altsum=0} relies on the
\textbf{telescoping sum principle} (or, for short, the \textbf{telescope
principle}):

\begin{theorem}
[telescoping sum principle]\label{thm.telescope.1}Let $u$ and $v$ be two
integers with $u\leq v+1$. Let $a_{u},a_{u+1},\ldots,a_{v},a_{v+1}$ be any
numbers. Then,%
\[
\sum_{j=u}^{v}\left(  a_{j+1}-a_{j}\right)  =a_{v+1}-a_{u}.
\]

\end{theorem}

\begin{proof}
First of all, the claim is obvious when $u=v+1$, because both sides are $0$ in
this case. Thus, WLOG assume that $u<v+1$, so the sum on the left is nonempty.
Then,\footnote{This is slightly informal. See the 2019 notes (\S 2.1.1) for a
formalization of this argument, as well as for a different proof.}%
\begin{align*}
&  \sum_{j=u}^{v}\left(  a_{j+1}-a_{j}\right) \\
&  =\left(  a_{u+1}-a_{u}\right)  +\left(  a_{u+2}-a_{u+1}\right)  +\left(
a_{u+3}-a_{u+2}\right)  +\cdots+\left(  a_{v+1}-a_{v}\right) \\
&  =\left(  a_{v+1}-a_{v}\right)  +\left(  a_{v}-a_{v-1}\right)  +\left(
a_{v-1}-a_{v-2}\right)  +\left(  a_{v-2}-a_{v-3}\right)  +\cdots+\left(
a_{u+1}-a_{u}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have flipped
the order of the summands}\right) \\
&  =a_{v+1}+\underbrace{\left(  -a_{v}+a_{v}\right)  }_{=0}%
+\underbrace{\left(  -a_{v-1}+a_{v-1}\right)  }_{=0}+\underbrace{\left(
-a_{v-2}+a_{v-2}\right)  }_{=0}+\cdots+\underbrace{\left(  -a_{u+1}%
-a_{u+1}\right)  }_{=0}-a_{u}\\
&  =a_{v+1}-a_{u}.
\end{align*}
This computation (with the many cancellations involved in it) is reminiscent
of a telescope being contracted; thus the name of the theorem.
\end{proof}

Note that the telescoping sum principle appears in many equivalent forms in
the literature, e.g., in the forms%
\begin{align*}
\sum_{j=u}^{v}\left(  a_{j}-a_{j-1}\right)   &  =a_{v}-a_{u-1}%
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\sum_{j=u}^{v}\left(  a_{j}-a_{j+1}\right)   &  =a_{u}-a_{v+1}.
\end{align*}
All such forms can be easily derived from Theorem \ref{thm.telescope.1} (e.g.,
by substituting $a_{j-1}$ for $a_{j}$, or by substituting $-a_{j}$ for $a_{j}%
$); alternatively, they can all be proved in similar ways.

Note also the similarity between Theorem \ref{thm.telescope.1} and the
Fundamental Theorem of Calculus:%
\[
\int_{u}^{v}F^{\prime}\left(  x\right)  dx=F\left(  v\right)  -F\left(
u\right)  .
\]
In a way, Theorem \ref{thm.telescope.1} is a discrete form of the Fundamental
Theorem of Calculus.

\bigskip

Now, we can reprove Proposition \ref{prop.binom.altsum=0}:

\begin{proof}
[Proof of Proposition \ref{prop.binom.altsum=0}.]The claim of Proposition
\ref{prop.binom.altsum=0} is clear if $n=0$ (since both sides are equal to $1$
in this case). Thus, we WLOG assume that $n\neq0$. Then,%
\begin{align*}
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\underbrace{\dbinom{n}{k}}%
_{\substack{=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}\\\text{(by Pascal's
recurrence)}}}  &  =\sum_{k=0}^{n}\left(  -1\right)  ^{k}\left(  \dbinom
{n-1}{k-1}+\dbinom{n-1}{k}\right) \\
&  =\sum_{k=0}^{n}\left(  \underbrace{\left(  -1\right)  ^{k}}_{=-\left(
-1\right)  ^{k-1}}\dbinom{n-1}{k-1}+\left(  -1\right)  ^{k}\dbinom{n-1}%
{k}\right) \\
&  =\sum_{k=0}^{n}\left(  -\left(  -1\right)  ^{k-1}\dbinom{n-1}{k-1}+\left(
-1\right)  ^{k}\dbinom{n-1}{k}\right) \\
&  =\sum_{k=0}^{n}\left(  \underbrace{\left(  -1\right)  ^{k}\dbinom{n-1}{k}%
}_{\text{let's call this }a_{k}}-\underbrace{\left(  -1\right)  ^{k-1}%
\dbinom{n-1}{k-1}}_{\text{then this is }a_{k-1}}\right) \\
&  =\sum_{k=0}^{n}\left(  a_{k}-a_{k-1}\right) \\
&  =a_{n}-a_{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the telescoping sum
principle}\right) \\
&  =\underbrace{\dbinom{n-1}{n}}_{\substack{=0\\\text{(since }n-1\in
\mathbb{N}\\\text{and }n-1<n\text{)}}}-\underbrace{\dbinom{n-1}{-1}%
}_{\substack{=0\\\text{(by definition)}}}=0-0\\
&  =0=\left[  n=0\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
n\neq0\right)  .
\end{align*}
Thus, Proposition \ref{prop.binom.altsum=0} is proved again.
\end{proof}

\bigskip

Using this second proof, we can show a generalization of Proposition
\ref{prop.binom.altsum=0}:

\begin{exercise}
[\textquotedblleft reverse hockey-stick identity\textquotedblright]Let
$n\in\mathbb{R}$. Let $m\in\mathbb{N}$. Prove that%
\[
\sum_{k=0}^{m}\left(  -1\right)  ^{k}\dbinom{n}{k}=\left(  -1\right)
^{m}\dbinom{n-1}{m}.
\]

\end{exercise}

(This is Exercise 2.1.1 in the 2019 notes.)

\subsubsection{A war between the odd and the even (aka sign-reversing
involutions)}

Now comes the perhaps nicest proof of Proposition \ref{prop.binom.altsum=0}.

\begin{proof}
[Third proof of Proposition \ref{prop.binom.altsum=0}.]As before, we can
easily dispatch the case when $n=0$. Thus, we assume that $n\neq0$. Hence, the
claim that we must prove simplifies to\footnote{The question mark about the
equality sign signifies that this equality has yet to be proved.}%
\begin{equation}
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}\overset{?}{=}0.
\label{pf.prop.binom.altsum=0.3rd.goal1}%
\end{equation}
Can we prove this combinatorially? Not directly, because the
LHS\footnote{\textquotedblleft LHS\textquotedblright\ = \textquotedblleft left
hand side\textquotedblright.} includes negative numbers, and negative numbers
don't count anything. However, we can resolve this issue in the simplest
possible way: We can bring the negative numbers to the RHS. So, the equality
(\ref{pf.prop.binom.altsum=0.3rd.goal1}) (which we intend to prove) gets
rewritten as follows:%
\begin{equation}
\sum_{k\text{ even}}\dbinom{n}{k}\overset{?}{=}\sum_{k\text{ odd}}\dbinom
{n}{k} \label{pf.prop.binom.altsum=0.3rd.goal2}%
\end{equation}
(where the summation signs \textquotedblleft$\sum_{k\text{ even}}%
$\textquotedblright\ and \textquotedblleft$\sum_{k\text{ odd}}$%
\textquotedblright\ are shorthand for \textquotedblleft$\sum_{\substack{k\in
\mathbb{N};\\k\text{ is even}}}$\textquotedblright\ and \textquotedblleft%
$\sum_{\substack{k\in\mathbb{N};\\k\text{ is odd}}}$\textquotedblright,
respectively). Equivalently, it gets rewritten as\footnote{An
\textquotedblleft\textbf{even-size}\textquotedblright\ set means a finite set
of even size. An \textquotedblleft\textbf{odd-size}\textquotedblright\ set
means a finite set of odd size.}%
\[
\left(  \text{\# of even-size subsets of }\left[  n\right]  \right)
\overset{?}{=}\left(  \text{\# of odd-size subsets of }\left[  n\right]
\right)
\]
(since $\sum_{k\text{ even}}\dbinom{n}{k}$ is the \# of even-size subsets of
$\left[  n\right]  $\ \ \ \ \footnote{Here, we are using the combinatorial
interpretation of BCs.}, whereas $\sum_{k\text{ odd}}\dbinom{n}{k}$ is the \#
of odd-size subsets of $\left[  n\right]  $).

Now, how can we prove such an equality? One way would be to find a bijection
between $\left\{  \text{even-size subsets of }\left[  n\right]  \right\}  $
and $\left\{  \text{odd-size subsets of }\left[  n\right]  \right\}  $. Such a
bijection can indeed be found: The map%
\begin{align*}
\left\{  \text{even-size subsets of }\left[  n\right]  \right\}   &
\rightarrow\left\{  \text{odd-size subsets of }\left[  n\right]  \right\}  ,\\
A  &  \mapsto%
\begin{cases}
A\setminus\left\{  1\right\}  , & \text{if }1\in A;\\
A\cup\left\{  1\right\}  , & \text{if }1\notin A
\end{cases}
\end{align*}
(that is, the map that removes the element $1$ from any subset that contains
$1$, and inserts $1$ into any subset that does not)\footnote{For example, this
map sends $\left\{  1,4,5,7\right\}  $ to $\left\{  4,5,7\right\}  $, and
sends $\left\{  2,4,5\right\}  $ to $\left\{  1,2,4,5\right\}  $.} is
well-defined\footnote{Indeed, it changes the size of any subset by $1$ upwards
or downwards, so an even-size subset will always turn into an odd-size
subset.} and bijective\footnote{Indeed, its inverse is the map%
\begin{align*}
\left\{  \text{odd-size subsets of }\left[  n\right]  \right\}   &
\rightarrow\left\{  \text{even-size subsets of }\left[  n\right]  \right\}
,\\
A  &  \mapsto%
\begin{cases}
A\setminus\left\{  1\right\}  , & \text{if }1\in A;\\
A\cup\left\{  1\right\}  , & \text{if }1\notin A
\end{cases}
\end{align*}
(given by the exact same formula!), because:
\par
\begin{itemize}
\item If a subset $A$ of $\left[  n\right]  $ contains $1$, then $\left(
A\setminus\left\{  1\right\}  \right)  \cup\left\{  1\right\}  =A$.
\par
\item If a subset $A$ of $\left[  n\right]  $ does not contain $1$, then
$\left(  A\cup\left\{  1\right\}  \right)  \setminus\left\{  1\right\}  =A$.
\end{itemize}
}. Thus, this map is a bijection. Hence, the bijection principle yields that
\[
\left(  \text{\# of even-size subsets of }\left[  n\right]  \right)  =\left(
\text{\# of odd-size subsets of }\left[  n\right]  \right)  .
\]
As explained above, this equality is just a rewritten form of
(\ref{pf.prop.binom.altsum=0.3rd.goal2}), which in turn is equivalent to
(\ref{pf.prop.binom.altsum=0.3rd.goal1}). Thus,
(\ref{pf.prop.binom.altsum=0.3rd.goal1}) holds, and Proposition
\ref{prop.binom.altsum=0} is proved (for the third time).
\end{proof}

The bijection that we used in the above proof is worth saying a few words
about. Let me drop the \textquotedblleft even-size\textquotedblright\ and
\textquotedblleft odd-size\textquotedblright\ conditions in its definition,
and thus extend it to a map%
\begin{align*}
t:\left\{  \text{subsets of }\left[  n\right]  \right\}   &  \rightarrow
\left\{  \text{subsets of }\left[  n\right]  \right\}  ,\\
A  &  \mapsto%
\begin{cases}
A\setminus\left\{  1\right\}  , & \text{if }1\in A;\\
A\cup\left\{  1\right\}  , & \text{if }1\notin A.
\end{cases}
\end{align*}
This map $t$ is called \textquotedblleft toggling the element $1$%
\textquotedblright, since the element $1$ is moved either into or out of the
subset. It is not just some bijective map, but actually a map that is inverse
to itself (i.e., it satisfies $t\circ t=\operatorname*{id}$); such maps are
known as \textbf{involutions}. It furthermore has the property that it flips
the sign $\left(  -1\right)  ^{\left\vert A\right\vert }$ (that is, it
satisfies $\left(  -1\right)  ^{\left\vert t\left(  A\right)  \right\vert
}=-\left(  -1\right)  ^{\left\vert A\right\vert }$ for every subset $A$ of
$\left[  n\right]  $). This makes it a \textquotedblleft\textbf{sign-reversing
involution}\textquotedblright. This sign-reversing property makes it suited
for canceling addends in alternating sums (i.e., sums in which different
addends can have different signs); many alternating-sum identities in
combinatorics can be proved using sign-reversing involutions. So the presence
of powers of $-1$ should not scare you away from using combinatorial arguments!

\bigskip

One last remark about the toggling involution $t$: It is a particular case of
a set-theoretical operation called \textbf{symmetric difference}. Here its definition:

\begin{definition}
Let $X$ and $Y$ be two sets. Then, their \textbf{symmetric difference}
$X\bigtriangleup Y$ is defined by%
\begin{align*}
X\bigtriangleup Y:=  &  \ \left\{  \text{all elements that belong to }X\text{
or }Y\text{ but not to both}\right\} \\
=  &  \ \left(  X\cup Y\right)  \setminus\left(  X\cap Y\right)  =\left(
X\setminus Y\right)  \cup\left(  Y\setminus X\right)  .
\end{align*}
In terms of Venn diagrams, $X\bigtriangleup Y$ is the grey zone in the
following Venn diagram (where the two circles are $X$ and $Y$):%
\[%
%TCIMACRO{\TeXButton{tikz venn diagram}{\begin{tikzpicture}
%\fill[gray, even odd rule] (0,0) circle (2) (1.5,0) circle (1.7);
%\end{tikzpicture}}}%
%BeginExpansion
\begin{tikzpicture}
\fill[gray, even odd rule] (0,0) circle (2) (1.5,0) circle (1.7);
\end{tikzpicture}%
%EndExpansion
\]

\end{definition}

Thus, the definition of our map $t$ can be rewritten as follows:%
\begin{align*}
t:\left\{  \text{subsets of }\left[  n\right]  \right\}   &  \rightarrow
\left\{  \text{subsets of }\left[  n\right]  \right\}  ,\\
A  &  \mapsto A\bigtriangleup\left\{  1\right\}  .
\end{align*}


See the 2019 notes (\S 2.1.2) for some properties of the symmetric difference,
particularly its associative law%
\[
\left(  A\bigtriangleup B\right)  \bigtriangleup C=A\bigtriangleup\left(
B\bigtriangleup C\right)  .
\]

\end{document}