\documentclass[11pt]{article}
\usepackage[margin=1in,headheight=24pt]{geometry}
\usepackage{fancyhdr}
\setlength{\headheight}{55pt}
\usepackage{hyperref}
\usepackage{tcolorbox}
\usepackage{xcolor}
\usepackage{amsfonts,amsmath,amssymb,amsthm}
\usepackage{mathtools}
\usepackage{subcaption}
\usepackage{setspace}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
\usepackage{tikz-network}

% Set line spacing to be more open
\setstretch{1.3}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}

\definecolor{black}{RGB}{0,0,0}
\definecolor{orange}{RGB}{230,159,0}
\definecolor{skyblue}{RGB}{86,180,233}
\definecolor{bluishgreen}{RGB}{0,158,115}
\definecolor{yellow}{RGB}{240,228,66}
\definecolor{blue}{RGB}{0,114,178}
\definecolor{vermillion}{RGB}{213,94,0}
\definecolor{reddishpurple}{RGB}{204,121,167}
\definecolor{cugold}{RGB}{207,184,124}

\pagestyle{plain}

\fancypagestyle{firstpage}{
  \fancyhf{}
  \renewcommand{\headrulewidth}{0pt}
  \fancyhead[c]{
    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied Combinatorics and Graph Theory} \hfill CU Denver} \\
    \rule{\textwidth}{0.5pt} \\
    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
  }
  \fancyfoot[C]{\thepage}
}

\newcommand{\scribebox}[4]{
\begin{tcolorbox}[colback=cugold!40,colframe=black,left=6pt,right=6pt,top=10pt,bottom=10pt]
    \centering
    \textbf{Lecture #1:} #2 \\
    \textbf{Date:} #3 \hfill      \textbf{Scribe:} #4
\end{tcolorbox}
}

%%% -+-+-+-+-+-+- BEGIN HERE -+-+-+-+-+-+- %%%
\newcommand{\lecturenumber}{$6$}
\newcommand{\lecturetitle}{Permutations, sets, and multisets}
\newcommand{\scribename}{Parsa S. Farahani}
\newcommand{\lecturedate}{February 9, 2026}

\begin{document}

\thispagestyle{firstpage}
\scribebox{\lecturenumber}{\lecturetitle}{\lecturedate}{\scribename}

\section{Permutations and factorials}

You may have seen this before: 
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] 

\vspace{0.5em}
\noindent The first goal of today is to unpack this (for the case where $n \ge 0$). \\[\baselineskip]
$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$ ($n$ factorial) \\[\baselineskip]
$n$ factorial is the number of different ways of ordering $[n]$ \\
(number of permutations of $[n]$) 

\begin{itemize}
    \item You have $n$ choices for which element of $[n]$ comes first.
    \item Given choice of first element, you have $n-1$ choices for which element comes second.
    \item Given choices of first and second, you have $n-2$ choices for third.
    \item And so on.
\end{itemize}

\vspace{1em}
Now, we want to understand $\binom{n}{k}$, i.e., ``$n$ choose $k$'' as the number of different ways of picking a subset of size $k$ from a set of size $n$. \\[\baselineskip]
define: 
\[ \binom{[n]}{k} := \{ T \subseteq [n] : |T| = k \} \] 
what we mean with ``$n$ choose $k$'' is:
\[ \binom{n}{k} = \left| \binom{[n]}{k} \right|. \]

\vspace{11 em}


\noindent Let $P([n], k)$ be the set of words of length $k$ using letters from $[n]$ without repetition. \\[\baselineskip]
- $|P([n], n)| = n!$ \\[\baselineskip]
- $|P([n], k)| = \underbrace{n \times (n-1) \times \dots \times (n-k+1)}_{k \text{ terms}}$ \\[\baselineskip]
denoted $n_{\downarrow k} = \frac{n!}{(n-k)!}
$ \\[\baselineskip]
$\Rightarrow n_{\downarrow n} = n!$

\vspace{1em}
\noindent Now, note that:
\[ \left| \binom{[n]}{k} \right| k! = |P([n], k)|. \] 
Since each set $T \in \binom{[n]}{k}$ can be ordered in $k!$ different ways. \\[\baselineskip]
$\Rightarrow \binom{n}{k} = \left| \binom{[n]}{k} \right| = \frac{|P([n], k)|}{k!} = \frac{n!}{(n-k)! k!}$ 

\vspace{1em}
By convention: 
\begin{itemize}
    \item $\binom{n}{k} = 0$ if $k < 0$ or $k > n$
    \item $\binom{n}{0} = 1$
    \item $\binom{0}{0} = 1$
\end{itemize}

\vspace{1.5em}

\section{Some identities}

\begin{lemma}
$\sum_{k} \binom{n}{k} = 2^n$
\end{lemma}
\begin{proof}
First, there are $2^n$ subsets of $[n]$. (recall product rule: $\underbrace{2 \times 2 \times \dots \times 2}_{n \text{ times}}$) \\[\baselineskip]
Also, a subset of $[n]$ must have some size $k$. \\
The sizes are mutually exclusive and collectively exhaustive. \\
Thus, use sum rule. 
\end{proof}

\vspace{4em}

\begin{lemma}
$\binom{n}{k} = \binom{n}{n-k}$
\end{lemma}
\begin{proof}
Let
\[
f :
\underbrace{2^{[n]}}_{\text{power set of }[n]} 
 \rightarrow 2^{[n]},\] \\[\baselineskip]
where for $S \in \binom{[n]}{k}$, we let $f(S) = [n] \setminus S $. \\[\baselineskip]
Here, $[n] \setminus S$ denotes the set of all elements of $[n]$ that are **not in $S$**, and since $S$ has $k$ elements, $[n] \setminus S$ has $n-k$ elements.
\vspace{1em}

\noindent Note: \\
$f \circ f = \text{Id}_f \Rightarrow f$ is its own inverse --- ``an involution'' \\[\baselineskip]
So $f$ restricts to a bijection between $\binom{[n]}{k}$ and $\binom{[n]}{n-k}$. \\[\baselineskip]
$\binom{n}{k} = \left| \binom{[n]}{k} \right| = \left| \binom{[n]}{n-k} \right| = \binom{n}{n-k}$
\end{proof}

\vspace{1em}

\begin{lemma}
For $n \ge 1$, $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$.
\end{lemma}
\begin{proof}
Let
\[
\mathcal{S}_1 = \{ S \in \binom{[n]}{k} : n \in S \},
\qquad
\mathcal{S}_2 = \{ S \in \binom{[n]}{k} : n \notin S \}.
\]
Then
\[
\binom{[n]}{k} = \mathcal{S}_1 \sqcup \mathcal{S}_2,
\]
where $\sqcup$ denotes disjoint union. \\[\baselineskip]
By sum rule, 
\[ \binom{n}{k} = \left| \binom{[n]}{k} \right| = |\mathcal{S}_1| + |\mathcal{S}_2|. \] 
If $S \in \mathcal{S}_1 \Rightarrow S - n \in \binom{[n-1]}{k-1}$. \\
$|\mathcal{S}_1| = \binom{n-1}{k-1}$ \\[\baselineskip]
If $S \in \mathcal{S}_2 \xrightarrow{\text{bijection}} S \in \binom{[n-1]}{k}$. \\
$|\mathcal{S}_2| = \left| \binom{[n-1]}{k} \right| = \binom{n-1}{k}$ \\[\baselineskip]
we plug these into $|\mathcal{S}_1| + |\mathcal{S}_2|$. 
\end{proof}

\vspace{1.5em}

\section{Multisets}

Multisets are unordered sets with repetitions accounted for. \\[\baselineskip]
For example, 
\[ \{a, a, b, b, c\} = \{a, b, a, b, c\} \text{ as a multiset}. \] 
But 
\[ \{a, a, b, b, c\} \neq \{a, b, c\} \text{ as a multiset}. \] 

\vspace{1em}
Let $\left(\!\binom{[n]}{k}\!\right)$ as the set of multisets on $[n]$ of size $k$. 
\[ \left(\!\binom{n}{k}\!\right) = \left| \left(\!\binom{[n]}{k}\!\right) \right|. \] 

\begin{example}
$\left(\!\binom{[3]}{2}\!\right) = \{ \{1,1\}, \{1,2\}, \{1,3\}, \{2,2\}, \{2,3\}, \{3,3\} \}$ \\
$\left(\!\binom{3}{2}\!\right) = \left| \left(\!\binom{[3]}{2}\!\right) \right| = 6$
\end{example}

\vspace{1.5em}

\section{Dots and bars}

\begin{lemma}
For $n, k \ge 0$, $\left(\!\binom{n}{k}\!\right) = \binom{n+k-1}{k}$.
\end{lemma}
\begin{proof}
We need to pick $k$ things from $n$ different kinds of things. \\[\baselineskip]
Form a list of ``dots and bars''. \\
We need $k$ dots and $n-1$ bars. \\
This list will fully determine how many things of each kind we pick. \\[\baselineskip]
For example: $\left(\!\binom{3}{5}\!\right)$: \\
1) $\bullet \bullet \mid \mid \bullet \bullet \bullet \Rightarrow$ 2 of first kind, 0 of second kind, 3 of third kind. \\
2) $\mid \bullet \bullet \bullet \mid \bullet \bullet \Rightarrow$ 0 of first kind, 3 of second kind, 2 of third kind. \\[\baselineskip]
Our list of dots and bars is of length $k + n - 1 = n + k - 1$. \\[\baselineskip]
We know that $k$ of the elements of the list must be dots (the rest are all bars). \\[\baselineskip]
How many different options for this choice? $\binom{n+k-1}{k}$. 
\end{proof}

\vspace{8 em}

\section{Lattice paths}



\begin{figure}[h]
    \centering
    \caption*{(Cartesian plane and its integer points)}
    \begin{tikzpicture}[scale=0.7]
        \draw[step=1cm,gray!30,very thin] (-0.5,-0.5) grid (4.5,3.5);
        \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};
        \draw[thick,->] (0,-0.5) -- (0,4) node[above] {$y$};
        \foreach \x in {0,1,2,3,4}
            \foreach \y in {0,1,2,3}
                \fill[red!60] (\x,\y) circle (2pt);
        \draw[ultra thick, blue] (0,0) -- (2,0) -- (2,1) -- (4,1) -- (4,3);
        \node[below left] at (0,0) {$(0,0)$};
        \node[above right] at (4,3) {$(m,n)$};
    \end{tikzpicture}
    \caption{A lattice path from $(0,0)$ to $(m,n)$.}
\end{figure}


A lattice path from $(0,0)$ to $(m,n)$ is a path from $(0,0)$ to $(m,n)$ using only lattice moves allowed: 
\begin{itemize}
    \item North
    \item East
\end{itemize}

\vspace{1em}
\begin{lemma}
The number of N,E lattice paths from $(0,0)$ to $(m,n)$ is $\binom{m+n}{m} = \binom{m+n}{n}$.
\end{lemma}
\begin{proof}
These paths are a list of size $m+n$ with $m$ elements being $E$ steps and $n$ elements being $N$ steps. \\[\baselineskip]
The choice of $E$ steps fully determines the path. \\[\baselineskip]
There are $\binom{m+n}{m}$ choices. Identity is analogous.
\end{proof}

\end{document}