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\begin{document}
\section*{Math 222 Fall 2022, Lecture 8: 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}{2}

\subsection{Factorials and binomial coefficients (cont'd)}

\setcounter{subsubsection}{6}

\subsubsection{Other properties of binomial coefficients (cont'd)}

\setcounter{theo}{26}

Here is another curious property of the $n$-th row of Pascal's triangle:

\begin{proposition}
Let $n$ be a positive integer. Then,%
\[
\dbinom{n}{0}+\dbinom{n}{2}+\dbinom{n}{4}+\cdots=\dbinom{n}{1}+\dbinom{n}%
{3}+\dbinom{n}{5}+\cdots=2^{n-1}.
\]

\end{proposition}

\begin{proof}
Let%
\begin{align*}
e  &  :=\dbinom{n}{0}+\dbinom{n}{2}+\dbinom{n}{4}+\cdots
\ \ \ \ \ \ \ \ \ \ \text{and}\\
o  &  :=\dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}+\cdots.
\end{align*}
(The letters stand for \textquotedblleft even\textquotedblright\ and
\textquotedblleft odd\textquotedblright; no relation to $2.718\ldots$ and
$0$.) Then,%
\begin{align*}
e+o  &  =\left(  \dbinom{n}{0}+\dbinom{n}{2}+\dbinom{n}{4}+\cdots\right)
+\left(  \dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}+\cdots\right) \\
&  =\dbinom{n}{0}+\dbinom{n}{1}+\dbinom{n}{2}+\cdots=\sum_{k\in\mathbb{N}%
}\dbinom{n}{k}\\
&  =\sum_{k=0}^{n}\dbinom{n}{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\dbinom{n}{k}=0\text{ for all }k>n\right) \\
&  =2^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Corollary 1.3.22 in Lecture
7}\right)
\end{align*}
and%
\begin{align*}
e-o  &  =\left(  \dbinom{n}{0}+\dbinom{n}{2}+\dbinom{n}{4}+\cdots\right)
-\left(  \dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}+\cdots\right) \\
&  =\dbinom{n}{0}-\dbinom{n}{1}+\dbinom{n}{2}-\dbinom{n}{3}\pm\cdots
=\sum_{k\in\mathbb{N}}\left(  -1\right)  ^{k}\dbinom{n}{k}\\
&  =\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\dbinom{n}{k}=0\text{ for all
}k>n\right) \\
&  =\left[  n=0\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
1.3.23 in Lecture 7}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n\text{ is positive, so that
}n\text{ cannot be }0\right)  .
\end{align*}


These two equations can be solved for $e$ and $o$ (just like any solvable
system of $2$ linear equations in $2$ unknowns); the result is%
\[
e=o=\dfrac{2^{n}}{2}=2^{n-1}.
\]
This proves the proposition.
\end{proof}

\bigskip

The next property of BCs is highly useful:

\begin{proposition}
[trinomial revision formula]\label{prop.binom.tri-rev}Let $n,a,b\in\mathbb{R}%
$. Then,%
\[
\dbinom{n}{a}\dbinom{a}{b}=\dbinom{n}{b}\dbinom{n-b}{a-b}.
\]

\end{proposition}

\begin{proof}
Four cases are possible:

\textit{Case 1:} We have $b\notin\mathbb{N}$.

\textit{Case 2:} We have $b\in\mathbb{N}$ and $a\notin\mathbb{N}$.

\textit{Case 3:} We have $b\in\mathbb{N}$ and $a\in\mathbb{N}$ and $a<b$.

\textit{Case 4:} We have $b\in\mathbb{N}$ and $a\in\mathbb{N}$ and $a\geq b$.

In each of the Cases 1, 2 and 3, the claim boils down to $0=0$ (check
it!\footnote{Details can be found in \S 2.2.1 of the 2019 notes.}). So we only
need to consider Case 4. In this case,%
\begin{align*}
\dbinom{n}{a}  &  =\dfrac{n\left(  n-1\right)  \left(  n-2\right)
\cdots\left(  n-a+1\right)  }{a!};\\
\dbinom{a}{b}  &  =\dfrac{a!}{b!\left(  a-b\right)  !}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the factorial formula,}\\
\text{since }b\in\mathbb{N}\text{ and }a\in\mathbb{N}\text{ and }a\geq b
\end{array}
\right)  ;\\
\dbinom{n}{b}  &  =\dfrac{n\left(  n-1\right)  \left(  n-2\right)
\cdots\left(  n-b+1\right)  }{b!};\\
\dbinom{n-b}{a-b}  &  =\dfrac{\left(  n-b\right)  \left(  n-b-1\right)
\left(  n-b-2\right)  \cdots\left(  n-a+1\right)  }{\left(  a-b\right)  !}.
\end{align*}
So we only need to prove that%
\begin{align*}
&  \dfrac{n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(  n-a+1\right)
}{a!}\cdot\dfrac{a!}{b!\left(  a-b\right)  !}\\
&  =\dfrac{n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-b+1\right)  }{b!}\cdot\dfrac{\left(  n-b\right)  \left(  n-b-1\right)
\left(  n-b-2\right)  \cdots\left(  n-a+1\right)  }{\left(  a-b\right)  !}.
\end{align*}
Cancelling out all the factorials, we can simplify this to%
\begin{align*}
&  n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(  n-a+1\right) \\
&  =\left(  n\left(  n-1\right)  \left(  n-2\right)  \cdots\left(
n-b+1\right)  \right)  \cdot\left(  \left(  n-b\right)  \left(  n-b-1\right)
\left(  n-b-2\right)  \cdots\left(  n-a+1\right)  \right)  .
\end{align*}
But this is clear from basic properties of finite products, since $0\leq b\leq
a$. So the trinomial revision formula is proved.
\end{proof}

As a particular case of trinomial revision, we obtain the \textbf{absorption
formula}:

\begin{proposition}
[absorption formula]Let $n,m\in\mathbb{R}$ such that $n\neq0$. Then,%
\[
\dbinom{m}{n}=\dfrac{m}{n}\dbinom{m-1}{n-1}.
\]

\end{proposition}

\begin{proof}
Apply the trinomial revision formula to $m$, $n$ and $1$ instead of $n$, $a$
and $b$. Then, recall that $\dbinom{m}{1}=m$ and $\dbinom{n}{1}=n$.
\end{proof}

\bigskip

Here are some less trivial properties of BCs. Note that these are variations
(and in some sense, generalizations) of the binomial theorem.

\begin{theorem}
[Chu--Vandermonde identity, aka Vandermonde convolution]Let $n\in\mathbb{N}$,
and let $x,y\in\mathbb{R}$. Then,%
\[
\dbinom{x+y}{n}=\sum_{k=0}^{n}\dbinom{x}{k}\dbinom{y}{n-k}.
\]

\end{theorem}

\begin{theorem}
[Cauchy's binomial formula]Let $n\in\mathbb{N}$ and $x,y,z\in\mathbb{R}$.
Then,%
\[
\sum_{k=0}^{n}\dbinom{n}{k}\left(  x+kz\right)  ^{k}\left(  y-kz\right)
^{n-k}=\sum_{k=0}^{n}\dfrac{n!}{k!}\left(  x+y\right)  ^{k}z^{n-k}.
\]

\end{theorem}

\begin{theorem}
[Abel's binomial formula]Let $n\in\mathbb{N}$ and $x,y,z\in\mathbb{R}$. Then,%
\[
\sum_{k=0}^{n}\dbinom{n}{k}\underbrace{x\left(  x+kz\right)  ^{k-1}%
}_{\substack{\text{This should be}\\\text{read as }1\\\text{if }k=0}}\left(
y-kz\right)  ^{n-k}=\left(  x+y\right)  ^{n}.
\]

\end{theorem}

We will prove the first of these three theorems in Lecture 15; the other two
will not be proved in this course (but their proofs can be found in
\cite[Exercise 1]{18f-mt3s}\footnote{More precisely, \cite[Exercise
1]{18f-mt3s} proves them in the case when $z=1$. However, the general case can
be easily obtained from this case (just replace $x,y,z$ by $\dfrac{x}%
{z},\dfrac{y}{z},1$).}). Note that the original binomial formula can be
derived from any of Cauchy's and Abel's binomial formulas by setting $z=0$.

\bigskip

Identities that involve BCs (like the ones we have discussed above) are called
\textbf{binomial identities}. There are many of them; I've seen whole books
devoted to them. Here is another example of a binomial identity:

\begin{proposition}
\label{prop.binom.sumknk}Let $n\in\mathbb{N}$. Then,%
\[
\sum_{k=0}^{n}k\dbinom{n}{k}=n\cdot2^{n-1}.
\]

\end{proposition}

\begin{proof}
[First proof (sketched).]The absorption formula yields $\dbinom{n}{k}%
=\dfrac{n}{k}\dbinom{n-1}{k-1}$ for each $k\neq0$. Thus, $k\dbinom{n}%
{k}=n\dbinom{n-1}{k-1}$ for each $k\neq0$. This also holds for $k=0$ for
pretty trivial reasons. Hence, if $n>0$, then%
\[
\sum_{k=0}^{n}k\dbinom{n}{k}=\sum_{k=0}^{n}n\dbinom{n-1}{k-1}%
=n\underbrace{\sum_{k=0}^{n}\dbinom{n-1}{k-1}}_{\substack{=\sum\limits_{k=-1}%
^{n-1}\dbinom{n-1}{k}=\sum\limits_{k=0}^{n-1}\dbinom{n-1}{k}=2^{n-1}%
\\\text{(by Corollary 1.3.22 in}\\\text{Lecture 7, since }n>0\text{)}}%
}=n\cdot2^{n-1}.
\]
You can easily check that this also holds for $n=0$. Thus, the proposition is proved.
\end{proof}

\begin{proof}
[Second proof (sketched).]The binomial formula (applied to $y=1$) yields%
\[
\sum_{k=0}^{n}\dbinom{n}{k}x^{k}=\left(  x+1\right)  ^{n}%
\ \ \ \ \ \ \ \ \ \ \text{for all }x\in\mathbb{R}.
\]
Take the derivative with respect to $x$. This results in%
\[
\sum_{k=0}^{n}\dbinom{n}{k}\cdot kx^{k-1}=n\cdot\left(  x+1\right)  ^{n-1}.
\]
Evaluating this at $x=1$, we obtain%
\[
\sum_{k=0}^{n}\dbinom{n}{k}\cdot k=n\cdot\left(  1+1\right)  ^{n-1}%
=n\cdot2^{n-1},
\]
which is the claim of Proposition \ref{prop.binom.sumknk}.
\end{proof}

We will see yet another proof of Proposition \ref{prop.binom.sumknk}
eventually (Remark 1.6.5 in Lecture 11).

\bigskip

\subsection{Counting subsets}

\subsubsection{All subsets}

We have already seen that if $S$ is an $n$-element set, and $k\in\mathbb{R}$,
then%
\[
\left(  \text{\# of }k\text{-element subsets of }S\right)  =\dbinom{n}{k}.
\]
So much for counting subsets of a given size. But what about counting
\textbf{all} subsets? The answer is really simple:

\begin{theorem}
\label{thm.subsets.all}Let $S$ be an $n$-element set (where $n\in\mathbb{N}$).
Then,%
\[
\left(  \text{\# of subsets of }S\right)  =2^{n}.
\]

\end{theorem}

\begin{proof}
Analogous to the proof of
\[
\left(  \text{\# of }k\text{-element subsets of }S\right)  =\dbinom{n}{k}.
\]
This time, instead of Pascal's recurrence, we have to use the identity
$2^{n-1}+2^{n-1}=2^{n}$.
\end{proof}

We will soon see some more artful proofs of the theorem.

\subsubsection{Lacunar subsets}

Let us move on to counting not all, but only some subsets, based on some
conditions that we want them to satisfy. Here is a simple-sounding one:

\begin{definition}
A set $S$ of integers is said to be \textbf{lacunar} if it contains no two
consecutive integers (i.e., there is no integer $i$ such that both $i$ and
$i+1$ belong to $S$).
\end{definition}

For instance, the set $\left\{  1,5,7\right\}  $ is lacunar, whereas $\left\{
1,5,6\right\}  $ is not. Note that any $1$-element or $0$-element set is lacunar.

The lacunar subsets of $\left[  5\right]  $ are%
\begin{align*}
&  \varnothing,\ \ \ \ \ \ \ \ \ \ \left\{  1\right\}
,\ \ \ \ \ \ \ \ \ \ \left\{  2\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{
3\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{  4\right\}
,\ \ \ \ \ \ \ \ \ \ \left\{  5\right\}  ,\\
&  \left\{  1,3\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{  1,4\right\}
,\ \ \ \ \ \ \ \ \ \ \left\{  1,5\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{
2,4\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{  2,5\right\}  ,\\
&  \left\{  3,5\right\}  ,\ \ \ \ \ \ \ \ \ \ \left\{  1,3,5\right\}  .
\end{align*}


The word \textquotedblleft lacunar\textquotedblright\ comes from
\textquotedblleft lacuna\textquotedblright\ (= \textquotedblleft
gap\textquotedblright). The same concept is known under different names
(essentially, every author invents their own terminology). You can think of a
lacunar set as a set of integers that value their privacy.

We can now ask several natural questions:

\begin{question}
\label{quest.lacsets.abc}For given $n,k\in\mathbb{N}$:

\textbf{(a)} How many lacunar subsets does $\left[  n\right]  $ have?

\textbf{(b)} How many $k$-element lacunar subsets does $\left[  n\right]  $ have?

\textbf{(c)} What is the largest size of a lacunar subset of $\left[
n\right]  $ ?
\end{question}

Let us start with part \textbf{(c)}. The answer is easiest to state in terms
of ceilings:

\begin{definition}
Let $x\in\mathbb{R}$. Then:

\begin{itemize}
\item We let $\left\lfloor x\right\rfloor $ denote the largest integer $\leq
x$. This integer $\left\lfloor x\right\rfloor $ is called the \textbf{floor}
of $x$, or \textquotedblleft\textbf{rounding down}\textquotedblright\ $x$.

\item We let $\left\lceil x\right\rceil $ denote the smallest integer $\geq
x$. This integer $\left\lceil x\right\rceil $ is called the \textbf{ceiling}
of $x$, or \textquotedblleft\textbf{rounding up}\textquotedblright\ $x$.
\end{itemize}
\end{definition}

For example:%
\begin{align*}
\left\lfloor 3\right\rfloor  &  =\left\lceil 3\right\rceil =3;\\
\left\lfloor 3.7\right\rfloor  &  =3,\ \ \ \ \ \ \ \ \ \ \left\lceil
3.7\right\rceil =4;\\
\left\lfloor \sqrt{2}\right\rfloor  &  =1,\ \ \ \ \ \ \ \ \ \ \left\lceil
\sqrt{2}\right\rceil =2;\\
\left\lfloor -2.5\right\rfloor  &  =-3,\ \ \ \ \ \ \ \ \ \ \left\lceil
-2.5\right\rceil =-2.
\end{align*}


Now we can state the answer to Question \ref{quest.lacsets.abc} \textbf{(c)}:

\begin{proposition}
\label{prop.lacsets.maxsize}Let $n\in\mathbb{N}$. Then, the largest size of a
lacunar subset of $\left[  n\right]  $ is $\left\lceil n/2\right\rceil $.
\end{proposition}

Our proof of this will rely on the following basic fact:

\begin{theorem}
[difference rule]\label{thm.diff-rule}Let $A$ be a finite set. Let $B$ be a
subset of $A$. Then:

\textbf{(a)} We have $\left\vert A\setminus B\right\vert =\left\vert
A\right\vert -\left\vert B\right\vert $.

\textbf{(b)} We have $\left\vert B\right\vert \leq\left\vert A\right\vert $.

\textbf{(c)} If $\left\vert B\right\vert =\left\vert A\right\vert $, then
$B=A$.
\end{theorem}

This theorem is basic set theory, so we won't prove it. Part \textbf{(a)} is
known as the \emph{difference rule}.

\begin{proof}
[Proof of Proposition \ref{prop.lacsets.maxsize}.]The set%
\[
\left\{  \text{all odd elements of }\left[  n\right]  \right\}  =\left\{
1<3<5<\cdots<\left(  n\text{ or }n-1\right)  \right\}
\]
(the $<$-signs on the right hand side here are simply saying that the elements
are being listed in increasing order) is a lacunar subset of $\left[
n\right]  $ whose size is $\left\lceil n/2\right\rceil $. So the size
$\left\lceil n/2\right\rceil $ is reachable. It remains to show that this size
is maximal, i.e., that every lacunar subset of $\left[  n\right]  $ has size
$\leq\left\lceil n/2\right\rceil $.

So let $L$ be a lacunar subset of $\left[  n\right]  $. Let $L^{+}$ be its
\textquotedblleft shadow\textquotedblright, defined by%
\[
L^{+}:=\left\{  \ell+1\ \mid\ \ell\in L\right\}  .
\]
Note that $L^{+}$ is a subset of $\left\{  2,3,\ldots,n+1\right\}  $.
Therefore, both $L$ and $L^{+}$ are subsets of $\left[  n+1\right]  $. Hence,
$L\cup L^{+}\subseteq\left[  n+1\right]  $. Theorem \ref{thm.diff-rule}
\textbf{(b)} thus yields $\left\vert L\cup L^{+}\right\vert \leq\left\vert
\left[  n+1\right]  \right\vert =n+1$.

However, the two sets $L$ and $L^{+}$ are disjoint (because if they had an
element $k$ in common, then we would have $k\in L$ and $k-1\in L$ (since $k\in
L^{+}$), and thus the two consecutive integers $k-1$ and $k$ would both belong
to $L$, which would contradict the lacunarity of $L$). Hence, by the sum rule,
we obtain $\left\vert L\cup L^{+}\right\vert =\left\vert L\right\vert
+\left\vert L^{+}\right\vert $. However, $\left\vert L^{+}\right\vert
=\left\vert L\right\vert $ (by the bijection rule, because the map
$L\rightarrow L^{+},\ \ell\mapsto\ell+1$ is a bijection). Thus,%
\[
\left\vert L\cup L^{+}\right\vert =\left\vert L\right\vert
+\underbrace{\left\vert L^{+}\right\vert }_{=\left\vert L\right\vert
}=\left\vert L\right\vert +\left\vert L\right\vert =2\cdot\left\vert
L\right\vert ,
\]
so that%
\[
2\cdot\left\vert L\right\vert =\left\vert L\cup L^{+}\right\vert \leq n+1.
\]
Hence, $\left\vert L\right\vert \leq\dfrac{n+1}{2}$. How can we derive
$\left\vert L\right\vert \leq\left\lceil n/2\right\rceil $ from this?

One way is to distinguish between the cases \textquotedblleft$n$
even\textquotedblright\ and \textquotedblleft$n$ odd\textquotedblright, and
make a (very simple) computation in each case. Another way is to argue that%
\[
\left\vert L\right\vert \leq\dfrac{n+1}{2}=\underbrace{\dfrac{n}{2}}%
_{\leq\left\lceil \dfrac{n}{2}\right\rceil }+\underbrace{\dfrac{1}{2}}%
_{<1}<\left\lceil \dfrac{n}{2}\right\rceil +1,
\]
and therefore (since both $\left\vert L\right\vert $ and $\left\lceil
\dfrac{n}{2}\right\rceil +1$ are integers) we get\footnote{Here, we are using
the fact that if $x$ and $y$ are two integers satisfying $x<y$, then $x\leq
y-1$.}%
\[
\left\vert L\right\vert \leq\left(  \left\lceil \dfrac{n}{2}\right\rceil
+1\right)  -1=\left\lceil \dfrac{n}{2}\right\rceil .
\]
So we have shown that the lacunar subset $L$ has size $\leq\left\lceil
n/2\right\rceil $. This completes the proof of Proposition
\ref{prop.lacsets.maxsize}.
\end{proof}

\bigskip

We note in passing that Theorem \ref{thm.diff-rule} \textbf{(b)} also helps us
prove the identity $\sum_{k=0}^{n}\dbinom{n}{k}=2^{n}$ (which we derived from
the binomial formula in Lecture 7) combinatorially:

\begin{proof}
[Combinatorial proof of $\sum_{k=0}^{n}\dbinom{n}{k}=2^{n}$.]Let
$n\in\mathbb{N}$. Then, each subset $S$ of $\left[  n\right]  $ has size
$\left\vert S\right\vert \in\left\{  0,1,\ldots,n\right\}  $ (since Theorem
\ref{thm.diff-rule} \textbf{(b)} yields $\left\vert S\right\vert
\leq\left\vert \left[  n\right]  \right\vert =n$). Thus, by the sum rule, we
have%
\begin{align*}
&  \left(  \text{\# of all subsets }S\text{ of }\left[  n\right]  \right) \\
&  =\sum_{k=0}^{n}\underbrace{\left(  \text{\# of all subsets }S\text{ of
}\left[  n\right]  \text{ that have size }k\right)  }_{=\left(  \text{\# of
all }k\text{-element subsets of }\left[  n\right]  \right)  =\dbinom{n}{k}}\\
&  =\sum_{k=0}^{n}\dbinom{n}{k}.
\end{align*}
Hence, $\sum_{k=0}^{n}\dbinom{n}{k}=\left(  \text{\# of all subsets }S\text{
of }\left[  n\right]  \right)  =2^{n}$ (by Theorem \ref{thm.subsets.all}), qed.
\end{proof}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[18f-mt3s]{18f-mt3s}Darij Grinberg, \textit{UMN Fall 2018 Math 5705
midterm \#3 with solutions}, \url{http://www.cip.ifi.lmu.de/~grinberg/t/18f/mt3s.pdf}
\end{thebibliography}


\end{document}