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\hbox to 6.28in { {\bf Combinatorial Analyis
\hfill Fall 2010} }
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\hbox to 6.28in { {\Large \hfill Lecture #1: #2 \hfill} }
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\lecture{1}{August 18}{Neil Calkin}{Sarah Anderson and Shihwei Chao}
\begin{flushleft}
Texts: Flajolet and Sedgwick Analytical Combinatoriccs
Wilf Generating functionalogy
Office: O-115
Email: calkin@clemson.edu
phone: 864-656-3437
web: www.math.clemson.edu/\textasciitilde{}calkin/
\end{flushleft}
Sorts of questions we will be addressing:
Questions regarding ``Combinatorial objects'' generally finite objects constructed by putting together simpler pieces.
Example:
The set of all finite sets of objects, each of which is a finite subset of the positive integers.
$\left\{ \left\{ 1,2,7\right\} ,\left\{ 3,15,92\right\} ,\left\{ 3,15,92,93\right\} ,\left\{ 5\right\} \right\} $
Typical questions for this class of objects:
How many? Infinitely many.
Can we split them up so that we get infinitely many natural classes,
naturally parametrized so that ech class is finite?
We can begin by thinking about what is interesting in this example? The largest element appearing is 93.
How many examples are there of objects with largest element $\leq93$?
Every subset which appears is a subset of $\left\{ 1,2,\ldots,93\right\} $. There are $2^{93}$such possible subsets. Therefore, there are $2^{2^{93}}$ objects.
How many examples are there of objects with largest element $=93$?
There are $2^{2^{93}}-2^{2^{92}}$ objects.
Counting often involves counting solutions to equations.
Examples:
``Compositions of $n$ into $k$ parts'' $=$ ways of writing $n$ as sum of $k$ natural numbers.
Compositions of $5$ into $2$ parts are
\begin{center}
$5+0$
$4+1$
$3+2$
$2+3$
$1+4$
$0+5$
\end{center}
Can also be described as counting solutions to
\begin{center}
$x_{1}+x_{2}+\cdots+x_{k}=n$ where $x_{1},x_{2},\ldots,x_{k}\in\mathbb{N}$.
\end{center}
Note: Here $3+2$ is regarded as distinct from $2+3$.
``Partitions of $n$ into $k$ parts'' $=$ ways of writing $x_{1}+x_{2}+\cdots+x_{k}=n$ where $x_{1}\geq x_{2}\geq\cdots\geq x_{k}\geq1$.
Example:
\begin{align*}
7 &=4+1+1+1 \\
&=3+2+1+1 \\
&=2+2+2+1
\end{align*}
are all partitions of $7$ into exactly $4$ parts.
Exercise:
How many ways are there to parenthesize the product $a_{1}a_{2}\cdots a_{n}$?
Hint:
$a_{1}$: Not interesting
$a_{1}a_{2}$: Not interesting
$\left(a_{1}a_{2}\right)a_{3}$ and $a_{1}\left(a_{2}a_{3}\right)$: There are 2 ways.
$\left(\left(a_{1}a_{2}\right)a_{3}\right)a_{4}$, $\left(a_{1}\left(a_{2}a_{3}\right)\right)a_{4}$,
$a_{1}\left(a_{2}\left(a_{3}a_{4}\right)\right)$, $a_{1}\left(\left(a_{2}a_{3}\right)a_{4}\right)$
$+$ how many others for $n=4$?
How do we list all of these in a good fashion?
Monday:
How to obtain $f\left(n\right)=$ numbers of parenthesise of the product
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$a_{1}a_{2}\cdots a_{n}$for $n\leq20$.
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Do obtain!
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