\documentclass[12pt]{article}
\usepackage{fancyhdr,fancybox,amssymb,epsfig,amsmath,latexsym,bm}
\usepackage{enumerate}
\usepackage{tikz}
\usepackage[all]{xy}
\usetikzlibrary{arrows}
\usetikzlibrary{decorations.markings}

\tikzset{big arrow/.style={decoration={markings,mark=at position 0.6
      with {\arrow[scale=2]{>}}}, postaction={decorate}, shorten >=1pt}}

% Code to make an augmented matrix (Usage: \begin{bmatrix}[ccc|c])
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
  \hskip -\arraycolsep
  \let\@ifnextchar\new@ifnextchar
  \array{#1}}
\makeatother

\topmargin=-.5in
\headsep=0.2in
\oddsidemargin=0in
\textwidth=6.7in
\textheight=9.2in
\footskip=.5in

% Change this line to \def\bold{\mathbf} for ``upright boldface.''
\def\bold{\bm}

\newcommand{\F}{\mathbb{F}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\def\0{\bold{0}}
\def\A{\bold{A}}
\def\B{\bold{B}}
\def\C{\bold{C}}
\def\I{\bold{I}}
\def\M{\bold{M}}
\def\P{\bold{P}}
\def\Q{\bold{Q}}
\def\S{\mathcal{S}}
\def\a{\bold{a}}
\def\b{\bold{b}}
\def\c{\bold{c}}
\def\e{\bold{e}}
\def\p{\bold{p}}
\def\q{\bold{q}}
\def\t{\bold{t}}
\def\u{\bold{u}}
\def\v{\bold{v}}
\def\x{\bold{x}}
\def\<{\langle}
\def\>{\rangle}
\def\hat{\widehat}
\def\longto{\longrightarrow}

\DeclareMathOperator{\Span}{Span}
\DeclareMathOperator{\rank}{rank}
\newcommand{\vv}[2]{\begin{bmatrix} #1 \\ #2 \end{bmatrix}}
\newcommand{\vvv}[3]{\begin{bmatrix} #1 \\ #2 \\ #3 \end{bmatrix}}

\newcommand{\ds}{\displaystyle}

\renewcommand{\footrulewidth}{1pt}
\newenvironment{proof}[1][Proof.]{\smallskip\begin{trivlist}
    \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip}
\newenvironment{sol}[1][Solution.]{\smallskip\begin{trivlist}
    \item[\hskip \labelsep {\sffamily #1}]}{\qed\end{trivlist}\bigskip}
\newenvironment{fminipage}
{\setlength{\fboxsep}{15pt}\begin{Sbox}\begin{minipage}}
{\end{minipage}\end{Sbox}\fbox{\TheSbox}}
\newcommand{\qed}{\hfill \ensuremath{\Box}}
\newcommand{\disp}{\displaystyle}

\def\multiset#1#2{\ensuremath{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{#1}{#2}\right)\kern-.3em\right)}}
\def\multichoose#1#2{\ensuremath{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{#1}{#2}\right)\kern-.3em\right)}}

\graphicspath{{./}{figs/}}

\pagestyle{fancy} \lhead{{\sf Homework 4 $|$ Due October 3 (Friday)
}} \rhead{\thepage} \cfoot{{\sf Math 4110 $|$ Combinatorics
    $|$ Fall 2025 $|$ M.~Macauley}}


\begin{document}

$\;$

\noindent\textbf{Topics}: Generating functions, compositions, and partitions.


\begin{enumerate}

\item Use Google and/or ChatGPT to find a really fun application of how generating functions can solve a particular mathematical problem. Write up a description that demonstrates that you understand all of the details. \\
  
  
\item In class, we derived the generating function for $p_{\leq
  k}(k)$, the number of partitions of $n$ where each part is at most
  $k$. Let $c_{\leq k}(n)$ be the number of compositions where each
  part is at most $k$.
  \begin{enumerate}
  \item Show that $c_{\leq k}(n)$ is the coefficient of $x^n$ in 
    \[
    \sum_{n=0}^\infty(x+x^2+\cdots+x^{k-1})^n.
    \]
  \item Derive the following generating function:
    \[
    \sum_{n=0}^\infty c_{\leq k}(n)x^n=\frac{1-x}{1-2x+x^k}.
    \]
  \end{enumerate}
  

\item In class, we constructed a bijection
  \[
  \big\{\text{partitions of $n$ into distinct parts}\big\}\longto
  \big\{\text{partitions of $n$ into odd parts}\big\}.
  \]
  The proof involved writing each distinct part as $2^i\cdot(2j+1)$,
  i.e., a power of $2$ times an odd number. This map can be inverted
  using the binary representation of the multiplicity of each odd
  part. Is there a ``base-$3$'' analogue of this result? If so, what would it look like? If not, explain what fails and why. \\

  
  
\item In class, we used probability generating functions to prove that it is impossible to assign weights to two six-sided dice so that the probability distribution of their sum is uniform. The argument was that the product
\[
P(x)Q(x)=\frac{1}{11}\big(1+x+\cdots+x^{10}\big)=\frac{1}{11}\cdot\frac{x^{11}-1}{x-1}
\]
cannot have all ten non-real $11^{\rm th}$ roots of unit as zeros.
\begin{enumerate}
  \item Show how this argument carries over for $n$-sided dice, for all
    even $n\geq 4$.
  \item Try to adapt this argument for $n$-sided dice, where $n$ is odd. Does it work? The main difference is that $-1$ is also a $(2n-1)^{\rm th}$ root of unity in this case. 
\end{enumerate}




  
\end{enumerate}

\end{document}




%% Move to HW 5.
%%
\item Recall the \emph{fundamental bijection}
  \[
  \theta\colon\S_n\longto\S_n,\qquad\theta\colon\pi\longmapsto\widehat{\pi},
  \]
  that maps a permutation in its standard cycle notation to the
  permutation whose 1-line notation is the result of removing the
  parantheses.
  \begin{enumerate}
  \item Prove that the following are equivalent:
    \begin{enumerate}[(i)]
    \item $\widehat{\pi}=\pi$,
    \item $\pi^{-1}=\pi$,
    \item $\pi$ is a product of cycles of length at most $2$.
    \end{enumerate}
  \item Prove that the number of such permutations is a Fibonacci
    number. [\emph{Hint}: show that they satisfy the Fibonacci
      recurrence.]
  \end{enumerate}
  




