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\begin{document}
\section*{Math 222 Fall 2022, Lecture 14: Binomial coefficients}

\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}}}

\setcounter{section}{1}

\section{Binomial coefficients (cont'd)}

\setcounter{subsection}{1}

\subsection{Revisiting trinomial revision}

In Lecture 8, we have proved the following:

\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}

We proved this proposition algebraically. Let us now try to prove it combinatorially.

\subsubsection{A partial proof by double counting}

We first consider the case when $n\in\mathbb{N}$.

\begin{proof}
[Partial proof of Proposition \ref{prop.binom.tri-rev} in the case when
$n\in\mathbb{N}$.]Let us first argue informally.

Assume that $n\in\mathbb{N}$. Fix a set $N$ of $n$ people.

We consider pairs $\left(  A,B\right)  $, where $A$ is a committee consisting
of $a$ people from $N$ (that is, an $a$-element subset of $N$), and where $B$
is a subcommittee consisting of $b$ people from $A$ (that is, a $b$-element
subset of $A$). Such pairs will be called \textbf{CS pairs} (for
\textquotedblleft committee-subcommittee pairs\textquotedblright). Formally
speaking, a CS pair is a pair $\left(  A,B\right)  $ of sets satisfying%
\[
B\subseteq A\subseteq N\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left\vert A\right\vert =a\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left\vert B\right\vert =b.
\]
Here is a symbolic picture of this situation:%
\[
\begin{tikzpicture}
\draw[fill=yellow!50] (0.5, 0.5) rectangle +(6, 3);
\draw[fill=green!50] (3, 2) ellipse (2 and 1);
\draw[fill=red!50] (2, 2) ellipse (0.5 and 0.5);
\node at (2, 2) {$B$};
\node at (4, 2) {$A$};
\node at (6, 3) {$N$};
\end{tikzpicture}\ \ .
\]


How many CS pairs are there? Let us count this number in two ways:

\begin{itemize}
\item \textit{First way:} We construct a CS pair $\left(  A,B\right)  $ by
first choosing $A$ and then choosing $B$. We have $\dbinom{n}{a}$ many options
for $A$ (since $A$ has to be an $a$-element subset of the $n$-element set
$N$). Once $A$ is chosen, we then have $\dbinom{a}{b}$ many options for $B$
(since $B$ has to be a $b$-element subset of the $a$-element set $A$). So, in
total, there are $\dbinom{n}{a}\dbinom{a}{b}$ many ways to make these choices.

\item \textit{Second way:} We construct a CS pair $\left(  A,B\right)  $ by
first choosing $B$ and then choosing $A$. We have $\dbinom{n}{b}$ many options
for $B$. Once $B$ is chosen, we then have $\dbinom{n-b}{a-b}$ many options for
$A$ (by Proposition 1.4.15 on Lecture 10, since $A$ has to be a subset of $N$
that contains $B$ as a subset). So, in total, there are $\dbinom{n}{b}%
\dbinom{n-b}{a-b}$ many ways to make these choices.
\end{itemize}

But these two ways are counting the same number. Thus, by comparing them, we
obtain
\[
\dbinom{n}{a}\dbinom{a}{b}=\dbinom{n}{b}\dbinom{n-b}{a-b}.
\]
This proves Proposition \ref{prop.binom.tri-rev} again in the case when
$n\in\mathbb{N}$.
\end{proof}

How can the above argument be formalized? On the surface, it looks like we
have applied the product rule to get the products $\dbinom{n}{a}\dbinom{a}{b}$
and $\dbinom{n}{b}\dbinom{n-b}{a-b}$. However, this is not the case. To get a
product like $\dbinom{n}{a}\dbinom{a}{b}$ from the product rule, we'd have to
find two sets $X$ and $Y$ such that $\left\vert X\right\vert =\dbinom{n}{a}$
and $\left\vert Y\right\vert =\dbinom{a}{b}$, and somehow our CS pairs would
have to be in bijection with the elements of $X\times Y$. Finding such an $X$
is easy:
\[
X=\mathcal{P}_{a}\left(  N\right)  =\left\{  a\text{-element subsets of
}N\right\}  .
\]
However, what would $Y$ be? We are tempted to set%
\[
Y=\mathcal{P}_{b}\left(  A\right)  =\left\{  b\text{-element subsets of
}A\right\}  ,
\]
which is however not allowed, since there is no \textquotedblleft%
$A$\textquotedblright\ fixed at this point. In other words, when we are
choosing a CS pair $\left(  A,B\right)  $, the options available for $B$
depend on our choice of $A$, but this is not how a Cartesian product would
work (a Cartesian product $X\times Y$ contains all pairs $\left(  x,y\right)
$ with $x\in X$ and $y\in Y$; the available options for $y$ do not depend on
the $x$).

Fortunately, at least the \textbf{number} of possible choices for $B$ does not
depend on the choice of $A$; this number is always $\dbinom{a}{b}$. In such a
situation, the total \# of ways is still a product, but this no longer follows
from the product rule; instead, we need the following more flexible rule:

\begin{statement}
\textbf{\textquotedblleft Dependent product rule\textquotedblright\ for two
sets:} Consider a situation in which you have to make two choices
(sequentially). Assume that you have $a_{1}$ options available in choice $1$,
and then (after making choice $1$) you have $a_{2}$ options available in
choice $2$ (no matter which option you chose in choice $1$). Then, the total
\# of ways to make both choices is $a_{1}a_{2}$.
\end{statement}

This statement is still fairly informal, but it can be formalized. (See
Theorem 2.2.2 in the 2019 notes for a formalization.)

This \textquotedblleft dependent product rule\textquotedblright\ can also be
extended to $n$ choices:

\begin{statement}
\textbf{\textquotedblleft Dependent product rule\textquotedblright\ for }$n$
\textbf{sets:} Consider a situation in which you have to make $n$ choices
(sequentially). Assume that you have $a_{1}$ options available in choice $1$,
and then (after making choice $1$) you have $a_{2}$ options available in
choice $2$ (no matter which option you chose in choice $1$), and then (after
making choices $1$ and $2$) you have $a_{3}$ options in choice $3$, and so on.
Then, the total \# of ways to make all the choices is $a_{1}a_{2}\cdots a_{n}$.
\end{statement}

\bigskip

However, there is a different (in my opinion, cleaner) way to formalize the
above proof of Proposition \ref{prop.binom.tri-rev}. Instead of tweaking the
product rule to apply to our situation, we just use the sum rule instead:

\begin{proof}
[Partial proof of Proposition \ref{prop.binom.tri-rev}, formalized using the
sum rule.]Again, let $n\in\mathbb{N}$. Pick an $n$-element set $N$. Define a
\textbf{CS pair} to be a pair $\left(  A,B\right)  $ of sets satisfying%
\[
B\subseteq A\subseteq N\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left\vert A\right\vert =a\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left\vert B\right\vert =b.
\]
Then, we shall count the \# of CS pairs in two ways.

\begin{itemize}
\item \textit{First way:} The first entry of a CS pair is always a subset $A$
of $N$ satisfying $\left\vert A\right\vert =a$. Hence, by the sum rule, we
have%
\begin{align*}
\left(  \text{\# of CS pairs}\right)   &  =\sum_{\substack{A\subseteq
N;\\\left\vert A\right\vert =a}}\underbrace{\left(  \text{\# of CS pairs whose
first entry is }A\right)  }_{\substack{=\left(  \text{\# of }b\text{-element
subsets of }A\right)  \\\text{(by the bijection principle, because
the}\\\text{map from }\left\{  \text{CS pairs whose first entry is }A\right\}
\\\text{to }\left\{  b\text{-element subsets of }A\right\}  \\\text{that sends
each CS pair }\left(  A,B\right)  \text{ to }B\text{ is a bijection)}}}\\
&  =\sum_{\substack{A\subseteq N;\\\left\vert A\right\vert =a}%
}\underbrace{\left(  \text{\# of }b\text{-element subsets of }A\right)
}_{=\dbinom{a}{b}}\\
&  =\sum_{\substack{A\subseteq N;\\\left\vert A\right\vert =a}}\dbinom{a}%
{b}=\dbinom{n}{a}\dbinom{a}{b}%
\end{align*}
(since the sum has $\dbinom{n}{a}$ many addends).

\item \textit{Second way:} The second entry of a CS pair is always a subset
$B$ of $N$ satisfying $\left\vert B\right\vert =b$. Hence, by the sum rule, we
have%
\begin{align*}
\left(  \text{\# of CS pairs}\right)   &  =\sum_{\substack{B\subseteq
N;\\\left\vert B\right\vert =b}}\underbrace{\left(  \text{\# of CS pairs whose
second entry is }B\right)  }_{\substack{=\left(  \text{\# of }a\text{-element
sets }A\text{ such that }B\subseteq A\subseteq N\right)  \\\text{(by the
bijection principle, because the}\\\text{map from }\left\{  \text{CS pairs
whose second entry is }B\right\}  \\\text{to }\left\{  a\text{-element sets
}A\text{ such that }B\subseteq A\subseteq N\right\}  \\\text{that sends each
CS pair }\left(  A,B\right)  \text{ to }A\text{ is a bijection)}}}\\
&  =\sum_{\substack{B\subseteq N;\\\left\vert B\right\vert =b}%
}\underbrace{\left(  \text{\# of }a\text{-element sets }A\text{ such that
}B\subseteq A\subseteq N\right)  }_{\substack{=\dbinom{n-b}{a-b}\\\text{(by
Proposition 1.4.15 on Lecture 10)}}}\\
&  =\sum_{\substack{B\subseteq N;\\\left\vert B\right\vert =b}}\dbinom
{n-b}{a-b}=\dbinom{n}{b}\dbinom{n-b}{a-b}%
\end{align*}
(since the sum has $\dbinom{n}{b}$ many addends).
\end{itemize}

Comparing these results, we find $\dbinom{n}{a}\dbinom{a}{b}=\dbinom{n}%
{b}\dbinom{n-b}{a-b}$. This again proves Proposition \ref{prop.binom.tri-rev}
for $n\in\mathbb{N}$.
\end{proof}

What have we done here? We have used the sum rule instead of the
\textquotedblleft dependent product rule\textquotedblright\ in this
formalization. It boils down to the same thing because a sum of several equal
addends is just one of these addends times the number of the addends (i.e., we
have $\sum_{k\in K}\lambda=\left\vert K\right\vert \cdot\lambda$). However, in
general, the sum rule tends to be more versatile than the \textquotedblleft
dependent product rule\textquotedblright, since it also applies to situations
where the number of options in choice $2$ is not independent of choice $1$.
(Of course, this results in a sum which you may or may not be able to simplify.)

\bigskip

\subsubsection{The polynomial identity trick}

So we have found a nice combinatorial proof of Proposition
\ref{prop.binom.tri-rev}, which unfortunately works only for $n\in\mathbb{N}$.
This looks like a dead-end; after all, there is no way we can extend our
counting argument to $n=\dfrac{1}{2}$ or $n=-7$ or $n=\sqrt{2}$.

But is it? We will now learn a \textquotedblleft simple\textquotedblright%
\ trick that will help us extend this proof to the full generality of
Proposition \ref{prop.binom.tri-rev}. More generally, this trick will
guarantee that if a binomial identity of a certain kind (there are some
restrictions, which we will soon see, but the trinomial revision formula
qualifies) has been proved for all nonnegative integers, then it must
automatically hold for all real numbers (or complex numbers, or anything
else\footnote{An algebraically experienced reader will want to replace the
words \textquotedblleft anything else\textquotedblright\ by the less vague
\textquotedblleft elements of a commutative $\mathbb{Q}$%
-algebra\textquotedblright\ (or at least by \textquotedblleft elements of a
field of characteristic $0$\textquotedblright). This level of generality can
be useful, but there is nothing substantially new happening at it; every
argument we will be making for real numbers can be repeated verbatim for
commutative $\mathbb{Q}$-algebras or fields of characteristic $0$. We have
chosen real numbers just to have something definite to talk about.}). Thus,
when proving such identities, we will be able to restrict ourselves to
nonnegative integers without loss of generality.

This trick is called the \textbf{polynomial identity trick} (the simplest form
of what is called the \textbf{Zariski density argument} in algebraic
geometry). The underlying theorem is the following:

\begin{theorem}
[easy half of the Fundamental Theorem of Algebra]\label{thm.fta-easy}Let
$n\in\mathbb{N}$. Then, a nonzero polynomial of degree $\leq n$ has $\leq n$ roots.
\end{theorem}

Here (and in the rest of this section), \textquotedblleft
polynomial\textquotedblright\ means \textquotedblleft polynomial in $1$
indeterminate with real coefficients\textquotedblright, and we use the capital
letter \textquotedblleft$X$\textquotedblright\ to denote the indeterminate.

\begin{example}
\textbf{(a)} The polynomial $X^{3}-3X+1$ has $3$ real roots, and its degree is
$3$.

\textbf{(b)} The polynomial $X^{2}+2X+1=\left(  X+1\right)  ^{2}$ has only $1$
real root (unless you count with multiplicity), and its degree is $2$.

\textbf{(c)} The polynomial $X^{2}+1$ has $0$ real roots, and its degree is
$2$.

In all these cases, the \# of roots is $\leq$ to the degree, just as Theorem
\ref{thm.fta-easy} predicts.
\end{example}

\begin{proof}
[A few words about the proof of Theorem \ref{thm.fta-easy}.]The key lemma is
that if a polynomial $P$ has a root $a$, then $P$ is divisible by the linear
polynomial $X-a$. Thus, if a polynomial $P$ has several distinct roots
$a_{1},a_{2},\ldots,a_{k}$, then we can divide $P$ by $X-a_{1},\ X-a_{2}%
,\ \ldots,\ X-a_{k}$ in turn; this decreases its degree by $k$. Therefore, if
$P$ had degree $\leq n$ at the onset, then it will have degree $\leq n-k$ at
the end of this division process. But this degree cannot be negative (unless
$P=0$). So we conclude that $n-k\geq0$, and thus $k\leq n$. (For details, see
any textbook on abstract algebra. Some references can be found in \S 2.6.2 of
the 2019 notes.)
\end{proof}

We will mostly use Theorem \ref{thm.fta-easy} via the following two corollaries:

\begin{corollary}
\label{cor.fta-easy.c1}If a polynomial $P$ has infinitely many roots, then $P$
is the zero polynomial.
\end{corollary}

\begin{proof}
Assume the contrary. Thus, $P$ is nonzero. Hence, Theorem \ref{thm.fta-easy}
(applied to $n=\deg P$) yields that $P$ has $\leq n$ roots. This contradicts
the assumption that $P$ has infinitely many roots.
\end{proof}

\begin{corollary}
\label{cor.fta-easy.c2}Let $P$ and $Q$ be polynomials. Assume that%
\[
P\left(  x\right)  =Q\left(  x\right)  \ \ \ \ \ \ \ \ \ \ \text{for each
}x\in\mathbb{N}.
\]
Then, $P=Q$.
\end{corollary}

\begin{proof}
For each $x\in\mathbb{N}$, we have%
\[
\left(  P-Q\right)  \left(  x\right)  =P\left(  x\right)  -Q\left(  x\right)
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since we assumed that }P\left(  x\right)
=Q\left(  x\right)  \right)  .
\]
In other words, each $x\in\mathbb{N}$ is a root of the polynomial $P-Q$. So
the polynomial $P-Q$ has infinitely many roots, and thus (by Corollary
\ref{cor.fta-easy.c1}) is the zero polynomial. In other words, $P=Q$.
\end{proof}

Corollary \ref{cor.fta-easy.c2} allows us to prove a polynomial identity $P=Q$
by showing only that $P\left(  x\right)  =Q\left(  x\right)  $ for all
$x\in\mathbb{N}$. The latter equality can often be proved combinatorially.

\subsubsection{Salvaging our proof of trinomial revision}

Let us apply this technique to turn our above partial proof of Proposition
\ref{prop.binom.tri-rev} into a complete proof -- i.e., extending it from the
case $n\in\mathbb{N}$ to the case $n\in\mathbb{R}$. First, we explain in which
sense our trinomial revision formula%
\[
\dbinom{n}{a}\dbinom{a}{b}=\dbinom{n}{b}\dbinom{n-b}{a-b}%
\]
is a \textquotedblleft polynomial identity\textquotedblright. What makes this
work is the following:

\begin{definition}
\label{def.binom-pols}We have previously defined $\dbinom{n}{k}$ whenever $n$
and $k$ are numbers. We now extend this definition to the case when $n$ is a
polynomial (while $k$ is still a number).
\end{definition}

Thus, if $P$ is a polynomial\footnote{You may be more used to writing
$P\left(  X\right)  $ instead of $P$ when $P$ is a polynomial. There is
nothing wrong about this (it certainly helps one remember that the
indeterminate is called $X$), but it isn't strictly required. Writing
\textquotedblleft$P=2X^{2}+3X+7$\textquotedblright\ is just as good as writing
\textquotedblleft$P\left(  X\right)  =2X^{2}+3X+7$\textquotedblright.} and if
$k\in\mathbb{N}$, then%
\[
\dbinom{P}{k}=\dfrac{P\cdot\left(  P-1\right)  \cdot\left(  P-2\right)
\cdot\cdots\cdot\left(  P-k+1\right)  }{k!}%
\]
\footnote{I have written out the \textquotedblleft$\cdot$\textquotedblright%
\ signs so that the first product $P\cdot\left(  P-1\right)  $ is not mistaken
for the result of substituting $P-1$ into $P$.}. Note that the right hand side
here is clearly a polynomial (since a product of polynomials is a polynomial).
For example, for any $k\in\mathbb{N}$, we have%
\[
\dbinom{X}{k}=\dfrac{X\left(  X-1\right)  \left(  X-2\right)  \cdots\left(
X-k+1\right)  }{k!}.
\]
For instance,%
\[
\dbinom{X}{2}=\dfrac{X\left(  X-1\right)  }{2}=\dfrac{1}{2}X^{2}-\dfrac{1}%
{2}X.
\]


If $P$ is a polynomial and $k\notin\mathbb{N}$, then $\dbinom{P}{k}$ is again
a polynomial, although a pretty boring one (it is the zero polynomial, by definition).

Now, we can finish our second proof of Proposition \ref{prop.binom.tri-rev}:

\begin{proof}
[Second proof of Proposition \ref{prop.binom.tri-rev}: the last mile.]Fix
$a,b\in\mathbb{R}$. Consider the following two polynomials:%
\[
P:=\dbinom{X}{a}\dbinom{a}{b}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ Q:=\dbinom{X}{b}\dbinom{X-b}{a-b}.
\]
Having proved Proposition \ref{prop.binom.tri-rev} for the case $n\in
\mathbb{N}$, we thus know that $P\left(  n\right)  =Q\left(  n\right)  $ for
all $n\in\mathbb{N}$ (since this equation $P\left(  n\right)  =Q\left(
n\right)  $ is exactly equivalent to $\dbinom{n}{a}\dbinom{a}{b}=\dbinom{n}%
{b}\dbinom{n-b}{a-b}$). In other words, $P\left(  x\right)  =Q\left(
x\right)  $ for all $x\in\mathbb{N}$. Thus, by Corollary \ref{cor.fta-easy.c2}%
, we conclude that $P=Q$. By substituting $n$ for $X$ on both sides of this
equality $P=Q$, we conclude that%
\[
P\left(  n\right)  =Q\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \text{for any }%
n\in\mathbb{R}.
\]
In other words,%
\[
\dbinom{n}{a}\dbinom{a}{b}=\dbinom{n}{b}\dbinom{n-b}{a-b}%
\ \ \ \ \ \ \ \ \ \ \text{for any }n\in\mathbb{R}.
\]
Thus, Proposition \ref{prop.binom.tri-rev} holds for every $n\in\mathbb{R}$,
not just for $n\in\mathbb{N}$.
\end{proof}

The technique we have just applied to generalize our result from
$n\in\mathbb{N}$ to $n\in\mathbb{R}$ is known as the \textbf{polynomial
identity trick}. The following picture symbolically illustrates the rather
peculiar logical endgame that happened when we applied the trick:
\[
\hspace{-2pc}%
%TCIMACRO{\TeXButton{a proof by the polynomial identity trick}%
%{\begin{tikzpicture}
%\path(0, 0) node[ellipse, draw, x radius=4.5, y radius=1)](A) {$\dbinom{n}{a}
%\dbinom{a}{b} = \dbinom{n}{b} \dbinom{n-b}{a-b}\ $
%for all $n \in\mathbb{N}$};
%\path(4, 4.5) node[ellipse, draw, x radius=4, y radius=1)](B) {$\dbinom{X}{a}
%\dbinom{a}{b} = \dbinom{X}{b} \dbinom{X-b}{a-b}$};
%\path(7, 2) node[ellipse, draw, x radius=4.5, y radius=1)](C) {$\dbinom{n}{a}
%\dbinom{a}{b} = \dbinom{n}{b} \dbinom{n-b}{a-b}\ $
%for all $n \in\mathbb{R}$};
%\draw[double distance=3pt, thick, -{Implies}%
%, double] (A.north) -- (B.south west);
%\draw[double distance=3pt, thick, -{Implies}%
%, double] (B.south east) -- (C.north);
%\end{tikzpicture}}}%
%BeginExpansion
\begin{tikzpicture}
\path(0, 0) node[ellipse, draw, x radius=4.5, y radius=1)](A) {$\dbinom{n}{a}
\dbinom{a}{b} = \dbinom{n}{b} \dbinom{n-b}{a-b}\ $
for all $n \in\mathbb{N}$};
\path(4, 4.5) node[ellipse, draw, x radius=4, y radius=1)](B) {$\dbinom{X}{a}
\dbinom{a}{b} = \dbinom{X}{b} \dbinom{X-b}{a-b}$};
\path(7, 2) node[ellipse, draw, x radius=4.5, y radius=1)](C) {$\dbinom{n}{a}
\dbinom{a}{b} = \dbinom{n}{b} \dbinom{n-b}{a-b}\ $
for all $n \in\mathbb{R}$};
\draw[double distance=3pt, thick, -{Implies}%
, double] (A.north) -- (B.south west);
\draw[double distance=3pt, thick, -{Implies}%
, double] (B.south east) -- (C.north);
\end{tikzpicture}%
%EndExpansion
\]


We will soon see some more interesting applications of this technique.


\end{document}