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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
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    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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\newcommand{\lecturenumber}{$3$}
\newcommand{\lecturetitle}{Combinatorial Arguments}
\newcommand{\scribename}{Simon Ruland}
\newcommand{\lecturedate}{January 28, 2026}

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This lecture considers several strategies for constructing combinatorial proofs, including:
\begin{itemize}
    \item 
    ``Counting two ways"
    \item
    (Strong) induction
    \item
    The ``product principle" and the ``sum principle"
    \item 
    ``Extremality" and sequences on finite structures
\end{itemize}

\section{Counting Two Ways}

The idea of this technique is to show that two expressions are equal by showing they both count the same set.
As an example, consider the ``handshaking lemma":
\begin{theorem}
For any undirected graph $G = (V,E)$ we have that
\begin{equation*}
    \sum_{u \in V} \textrm{deg}(u) = 2|E|.
\end{equation*}
\end{theorem}
\begin{proof}
Let $P = \{(u,e)\in V \times E: u \in e\}$.
We first count this set as
\begin{equation*}
    |P| = \sum_{u\in V}|\{e \in E: u \in e\}| = \sum_{u\in V}\textrm{deg}(u).
\end{equation*}
We now count this set as
\begin{equation*}
    |P| = \sum_{e\in E} |\{u \in V: u \in e\}| = 2|E|,
\end{equation*}
where the last equality holds since every edge contains exactly two nodes.
\end{proof}

\section{Induction}

The idea of an inductive proof is to first directly show that the statement we are trying to prove is true for a ``small'' case (e.g., $n = 0$).
Then, we use what we assume (but soon will know) is true for ``smaller'' cases (e.g., for $1 < n \leq k$ for some $k$) to show that the statement is conditionally true for ``larger'' cases (e.g., $n = k +1)$.
The magic of induction is to transform our assumption to facts one step at a time given that we indeed started with a fact (e.g., which we directly showed for $n = 0$).

As an example of an inductive argument, we consider walks and paths on undirected graphs.
A walk of length $\ell$ is a sequence $v_0,v_1,v_2,\ldots,v_\ell \in V$ such that 
\begin{equation*}
    \{v_{i-1}, v_i\} \in E
\end{equation*}
for all $i \in [\ell]$.
In other words, a walk is a generalization of paths that allows repetitions.

\begin{theorem}
Let $G=(V,E)$ be an undirected graph. 
Then, any $\{s,t\}$-walk contains an $\{s,t\}$-path. 
\end{theorem}
\begin{proof}
Let $W$ be an $\{s,t\}$-walk $s = v_0,v_1,\ldots,v_\ell=t$.
For the base case, consider the case in which $\ell = 0$
Then, $W$ is automatically a path.

By way of induction, suppose that $\ell \geq 1$ and that every $\{s,t\}$-walk of length $0\leq k\leq \ell - 1$ contains an $\{s,t\}$-path.
We need to show that every $\{s,t\}$-walk of length $\ell$ contains an $\{s,t\}$-path.
If $W$ is a walk but not a path, then it contains a repeated node $v_i = v_j$ such that $i < j$.
Let $W' = v_0, v_1, \ldots, v_i, v_{j+1}, v_{j+2}, \ldots, v_\ell$. 
Then,  $W'$ is an $\{s,t\}$-walk of length strictly smaller than $\ell$.
By the inductive hypothesis, $W'$ contains an $\{s,t\}$-path. 
Therefore, so does $W$.
\end{proof}

\section{``Product Principle" and the ``Sum Principle"}

The idea of these proof techniques is that we can multiply the numbers of independent choices and sum over mutually exclusive and collectively exhaustive cases.

\begin{theorem}
Let $V = [n]$. 
Then,
\begin{enumerate}
    \item 
    There are $2^{\binom{n}{2}}$ different undirected graphs on $V$.
    \item 
    There are $\binom{\binom{n}{2}}{k}$ different undirected graphs on $V$ with $|E|=k$
    \item 
    $\sum_{k=0}^{\binom{n}{2}}\binom{\binom{n}{2}}{k} = 2^{\binom{n}{2}}$
\end{enumerate}
\end{theorem}
\begin{proof}
Since $V$ is given, a graph $G=(V,E)$ is completely determined by our choice of $E$.
We now prove each of the statements.
\begin{enumerate}
    \item 
    Note that
    $E\subseteq \{\{u,v\}\}: u\in V, v \in V\}$.
    That is, the set of candidate edges to construct our graph is of size $\binom{n}{2}$.
    Each edge can be included or excluded, so there are $2^{\binom{n}{2}}$ possible undirected graphs.
    \item 
    If $|E| = k$, then we must pick a set of size k from our set of candidates (without repetition). There are $\binom{\binom{n}{2}}{k}$ ways of doing this.
    \item
    Any graph on $G$ must have $0\leq k \leq \binom{n}{2}$; these options are mutually exclusive and collectively exhaustive.
    Therefore, we use the sum principle to obtain
    \begin{equation*}
    \sum_{k=0}^{\binom{n}{2}} \binom{\binom{n}{2}}{k} = 2^{\binom{n}{2}}
    \end{equation*}
\end{enumerate}
\end{proof}
More generally, for any $m \in \mathbb{N}$, we have
\begin{equation*}
\sum_{k=0}^{m} \binom{m}{k} = 2^m
\end{equation*}
This expression sums over the rows of Pascal's Triangle (check out \href{https://oeis.org/A007318}{OEIS A007318}).

\section{``Extremality" and Finite Structures}

\begin{theorem}
Let $D = (V,A)$ be a directed graph. 
Then, $\textrm{deg}^-(u) = \textrm{deg}^+(u) = 1$ for all $u \in V$ if and only if $D$ is the disjoint union of directed cycles.
\end{theorem}
\begin{proof}

Per usual, one direction is ``easy.''
If $D$ is the disjoint union of directed cycles, then each $u\in V$ is in a unique directed cycle as its entire component. 
So, $\text{deg}^-(u) = \text{deg}^+(u) = 1$ for all $u \in V$.

To prove the other direction, consider any $v_1 \in V$. 
Since $\textrm{deg}^+(v_1) = 1$, there is a unique $v_2 \in V$ such that $(v_1, v_2) \in A$. Similarly, there is a unique $v_3 \in V$ such that $(v_3, v_1) \in A$, and so on.
Inductively, there is a sequence $v_1,v_2,v_3,\ldots$ with corresponding arcs. 
Since $D$ is finite, at some point the sequence has a repetition. 
So, there exists a first (i.e., in this sense ``extreme'') pair $i<j$ with $v_i = v_j$.
Note that we must have $v_i = v_1$ since otherwise $(v_{i-1}, v_i) \in A$ and also $(v_{j-1}, v_j) \in A$; and these arcs are distinct by our choice of ``first'' repetition.
But then $\text{deg}^-(v_i)\geq2 $, which contradicts the assumption that $\textrm{deg}^-(u) = 1$ for all $u \in V$. 
So, $v_i=v_1$ and $v_1,v_2,...,v_{j-1},v_i=v_1$ forms a directed cycle. 
We can remove this cycle and further decompose the graph.
\end{proof}
\end{document}