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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}{The Principle of Inclusion-Exclusion (PIE) and Applications}
\newcommand{\scribename}{Simon Ruland}
\newcommand{\lecturedate}{February 25, 2026}

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Recall the sum principle:

$$|S \sqcup T| = |S| + |T|$$
The disjoint union is required.

You may recall from prior life that 
1) $$|S \cup T| = |S| + |T| - |S \cap T|$$

Every element in the intersection is counted twice in the first part. The double-counting must be corrected.

Goal: Generalize 1) to work with any finite collection

$$S_1, S_2, .. S_n \subseteq S$$

For n=2, $$|S| = |S_1 \cup S_2| + |S \setminus (S_1 \cup S_2)|$$

So if we know $|S|$ and $|S \setminus (S_1 \cup S_2)|$, then we know $|S_1 \cup S_2|$.
We will focus on terms like $|S \setminus (S_1 \cup S_2)|$.

\begin{theorem}
The Principle of Inclusion and Exclusion:
Let S be finite and $S_1, S_2, ..., S_n \subseteq S$.
Then, 
$$|S\setminus \bigcup_{1\leq i \leq n}S_i| = |S| - \sum_{1 \leq i \leq n} |S_i| + \sum_{1 \leq i < j \leq n} |S_i \cap S_j| - ... + (-1)^n |\bigcap_{1\leq i \leq n} S_i|$$
\end{theorem}

\begin{proof}
It suffices to show that for any s in S, its contribution to LHS is the same as its contribution to RHS.

Consider any s in S; there are two cases:

1) $$s\notin \bigcup_{i=1}S_i$$.

LHS: So, $s \in \bigcup_{i=1}^nS_i$ which contributes 1 to the count.

RHS: s is in S, so it is accounted for in $|S|$. s is not in any of the intersections, so in total it contributes 1 to the count of RHS.

2) $$s \in \bigcup_{i=1}^nS_i$$. Suppose $s \in S_i$ for exactly $m \geq 1$ of the sets $S_1, S_2, ...,S_n$

LHS: $$s \notin S\setminus U_{i=1}^nS_i$$
so it contributes 0 to LHS.

RHS: We know that $s \in S_i \cap S_{i_2}, ... S_{j_k}$, a k-subset of the m subsets that contain s. Based on the alternating signs,

$$0 = \binom{m}{0}-\binom{m}{1}+\binom{m}{2}- ... + (-1)^m\binom{m}{m}$$
because $$\sum_{\text{k is odd}}\binom{m}{k} = \sum_{\text{k is even}}\binom{m}{k}$$
\end{proof}

Application 1: Derangements

A permutation is a bijection

$$\pi:[n] \rightarrow [n] $$

For example, in one line notation and function notation we have

$$\pi = 4,1,2,3$$
$$\pi(1) = 2, \pi(2) = 3, \pi(3) = 4, \pi(4) = 1$$

A permutation is a derangement if $\pi(i) \neq i \ \forall i \in [n]$.

How many derangements of [n] are there?

Idea: use PIE to ``count with restrictions'':

\begin{enumerate}
\item S: the set of all objects

\item $S_i$: the subset of S that violates the ith restriction

\item $S \setminus\bigcup_{i=1}^n S_i$: the subset of S that meet all restrictions.
\end{enumerate}

In the case of derangements, 

\begin{enumerate}
\item $S=\mathcal{S}_n$, the set of all permutations on [n]
\item $S_i \subseteq \mathcal{S}_n$ such that $\pi \in S_i \iff \pi(i) = i$
\item $S\setminus\bigcup_{i=1}^n S_i$ is the set of derangements

So the number of derangements is $D(n) = |S\setminus\bigcup_{i=1}^n S_i|$
\end{enumerate}

$\pi = 1,2,...n \implies \pi \in S_i \forall i \in [n]$. The intersections over $S_1,..,S_n$ are non-trivial.

Note that any $\pi \in S_i$ looks like
$$\pi^{-1}(1), \pi^{-1}(2), ..., \pi^{-1}(i-1), i, \pi^{-1}(i+1), ...,  \pi^{-1}(n)$$
Except for the i, $\pi$ is a permutation on n-1 letters.

$|S_i| = (n-1)!$

There are $\binom{n}{1}$ choices for $S_i$, so 

$$ \sum_{i=1}^n |S_i| = \binom{n}{1}(n-1)! = n(n-1)! = n!$$

Similarly, any $\pi \in S_i \cap S_j$ looks like

$$\pi^{-1}(1), ..., \pi^{-1}(i-1), \pi^{-1}, i , (i+1), ..., \pi^{-1}(j-1), j, \pi^{-1}(j+1), ...,   \pi^{-1}(n)$$

Except for i and j, $\pi$ is a permutation on n-2 letters.

$$|S_i\cap S_j| = (n-2)!$$

There are $\binom{n}{2}$ options for $S_i, S_j$, so

$$\sum_{i<j}|S_i \cap S_j| = \binom{n}{2}(n-2)!$$

In general, for any k number of fixed points, the sum of k-term intersections:

$$\binom{n}{k}(n-k)! = \frac{n!}{k!}$$

To sum up (pun intended),

$$D(n) = |S\setminus \bigcup_{i=1}^nS_i| = n!-n!+\frac{n!}{2}-\frac{n!}{3}+...+(-1)^n\frac{n!}{n!}$$

$$D(n) = n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{(-1)^n}{n!}$$
For any $n\geq0$.

Corollary: $\lim_{n\to \infty}\frac{D(n)}{n!} = \frac{1}{e}$

\begin{proof}
For any fixed point n, $\frac{D(n)}{n!}$ is a truncation of the Taylor series approximation for $\frac{1}{e}$.
\end{proof}

So a random permutation has a 37 percent chance of being a derangement.

Application 2: Chromatic Polynomials

Given $G=(V,E)$

\begin{enumerate}
    \item S = $\{c:V \to [t]\}$
    \item For any $e=\{u,v\}, S_e \subseteq S$ such that $c(u)=c(v)$
    \item $S\setminus\bigcup_{e \in E} S_e$ is the set of proper colorings.
\end{enumerate}

We want $|S\setminus\bigcup_{e \in E} S_e|$

$|S| = t^n$ for n = $|V|$

Consider any $F \subseteq E$ and the induced subgraph $H_F = (V, F)$.

Note that any $c \in \bigcap_{e \in F}S_e$ makes all edges in $H_F$ monochromatic, so each connected component of $H_F$ is monochromatic.

$\kappa(H_F)$ is the number of connected components of $H_F$.

$|\bigcap_{e \in F}S_e| = t^{\kappa(H_F)}$ since the coloring of the different components are independent of one another.

Using PIE, we have that:
$$|S\setminus\bigcup_{e \in E} S_e| = \sum_{\emptyset \subseteq F \subseteq E} (-1)^{|F|}t^{\kappa(H_F)} = P(G;t)$$

Some observations: 

\begin{enumerate}
\item if $|F| = 0, \kappa(H_F)=n$.
\item if $|F|=1$, we are picking individual edges, so $\kappa(H_F) = n-1$ and the coefficient in front of $t^{n-1}$ is the number of edges $|E|$.
\item if $|F|=|E|$, then $\kappa(H_F=\kappa(G)$, so the smallest power of t with a nonzero coefficient is $t^\kappa(G)$.
\item P(G;t) is a polynomial.
\end{enumerate}
\end{document}