\documentclass[12pt]{article}
\usepackage{amsmath}  % this adds some math 
\usepackage{fullpage}
\usepackage{tikz}            % if you delete this, you will have trouble with the included picture. You can delete the picture.
\usepackage{chessboard}


% If you have a problem, please let me know sooner than later

\newcommand\name{Chuck Norris} %%%%%%%%%%%%%%  WRITE YOUR NAME HERE



\newcounter{excounter}
\setcounter{excounter}{1}
\newcommand\question[2]{\vskip 1em  \noindent\textbf{\arabic{excounter}\addtocounter{excounter}{1}:} \emph{#1} \noindent#2}


% You can also erase this if you do not have package hancyhdr
% Fancy footnote......... 
\usepackage{fancyhdr}  %% If it does not work with your latex installation, you may just delete this...
\pagestyle{fancy}
\usepackage{lastpage}
\rfoot{\name, page \thepage/\pageref{LastPage}}
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%%% Note that you do not have to use this template. You can use your own writeup or make any adjustment to this one you like.
%%% This is provided only for your convenience. 


\begin{document}


{\bf MATH304  \hspace{1cm} HW 6\hfill \name}
\vskip 1em

% left column
\begin{minipage}[t]{0.75\textwidth}
Due \textbf{Oct  8} before class, \textbf{answer without justification will receive 0 points}.
The solution has to be typed (using \LaTeX). 
\end{minipage}
% right box for score column
\begin{minipage}[t]{0.25\textwidth}
\hfill \begin{tabular}{|r|}
\hline
\hskip 3em Score \hskip 3em \\
\hline \\ \\
\hskip 3em /60 \\
\hline
\end{tabular}
\end{minipage}


%%%% Delete from here
Here is some extra \LaTeX{} code so you can create nicer math formulas.

Brackets, that have the  correct size. Use
\begin{verbatim}
\[
\left( \frac{4}{2} \right)   
+  \left(   \frac{ \left\{  4^{3^{7^8}}   \right] }{ 34} \right|
\]
\end{verbatim}
\[
\left( \frac{4}{2} \right) 
+  \left(   \frac{ \left\{  4^{3^{7^8}}   \right] }{ 34} \right|
\]

For aligning several equations use
\begin{verbatim}
\begin{align*}
  4 &= 6\\
  12 +12 +12 &> 13
\end{align*}
\end{verbatim}
\begin{align*}
  4 &= 6\\
  12 +12 +12 &> 13
\end{align*}


It is even possible to put equations side by side
\begin{verbatim}
\begin{align*}
  4 &= 6                         &     4+4 &\geq 4\\
  12 +12 +12 &> 13         &        9 &= 17292/30
\end{align*}
\end{verbatim}
\begin{align*}
  4 &= 6                         &     4+4 &\geq 4\\
  12 +12 +12 &> 13         &        9 &= 17292/30
\end{align*}


%%%% To here after you read it.


\question{}{
Prove that the only antichain of $S=\{1,2,3,4\}$ of size 6 is the antichain
of all 2-subsets of $S$. 
}

\textbf{Solution:}


\question{}{
Let $n$ and $k$ be a positive integers. Give a combinatorial proof that 
\[  \sum_{k=1}^nk\binom{n}{k}^2 = n \binom{2n-1}{n-1}.\]
}

\textbf{Solution:}



\question{}{
Let $n$ be a positive integer. Prove that
\[ \sum_{k=0}^n(-1)^k\binom{n}{k}^2 =  \begin{cases}  0 & \text{ if } n \text{ is odd } \\ 
(-1)^m\binom{2m}{m} & \text{ if } n=2m. \end{cases}  \]
\emph{Hint: consider $(1-x^2)^n = (1+x)^n(1-x)^n$.}
}

\textbf{Solution:}

\question{}{
Evaluate the sum
\[ 1 - \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} - \frac{1}{4}\binom{n}{3} + \cdots + (-1)^n \frac{1}{n+1}\binom{n}{n}.\]
}

\textbf{Solution:}



\question{}{
Use \textbf{combinatorial} reasoning to prove the identity (in the given form)
\[ \binom{n}{k} - \binom{n-3}{k} = \binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1} \]
}

\textbf{Solution:}


\question{}{
Use Newton's binomial theorem to approximate $\sqrt{80}$.\\
\emph{(Hint: See page 148 and 149. First three digits after the decimal point is enough.)}
}

\textbf{Solution:}



 
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