notes/school/di-ma/uebung/02/02_1.tex.old
Valentin Brandl e39bbdf827
Add dima u02
2018-10-27 16:57:32 +02:00

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\documentclass[12pt,a4paper,german]{article}
\usepackage{url}
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\usepackage{ngerman}
\usepackage{float}
\usepackage{diagbox}
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\usepackage{amsmath}
\usepackage{csquotes}
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\geometry{left=2.0cm,textwidth=17cm,top=3.5cm,textheight=23cm}
%%%%%%%%%% Fill out the the definitions %%%%%%%%%
\def \name {Valentin Brandl} %
\def \matrikel {108018274494} %
\def \pname {Marvin Herrmann} %
\def \pmatrikel {108018265436} %
\def \gruppe {2 (Mi 10-12 Andre)}
\def \uebung {2} %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DO NOT MODIFY THIS HEADER
\newcommand{\hwsol}{
\vspace*{-2cm}
\noindent \matrikel \quad \name \hfill \"Ubungsgruppe: \gruppe \\
\noindent \pmatrikel \quad \pname \\
\begin{center}{\Large \bf L\"osung f\"ur \"Ubung \# \uebung}\end{center}
}
\begin{document}
%Import header
\hwsol
\section*{Aufgabe 2.1}
\begin{enumerate}[1.]
\item Kekse $\widehat{=}$ Bälle (unterscheidbar, da \enquote{verschieden}), Portionen $\widehat{=}$ Urnen (nicht
unterscheidbar). $n = 9$, $k = 5$
Problem entspricht einer ungeordneten $k$-Mengenpartition, also $S_{n,k}$
\begin{eqnarray*}
S_{n,k} &=& S_{n-1,k-1} + k * S_{n-1,k} \text{ mit} \\
S_{0,0} &=& 1 \\
S_{n,n} &=& 1 \\
S_{n,1} &=& 1 \\
S_{n,2} &=& 2^{n-1} - 1 \\
S_{n,3} &=& \frac{1}{2}(3^{n-1} - 2^n + 1) \\
S_{n,0} &=& 0 \\\\
S_{9,5} &=& S_{8,4} + 5 * S_{8,5} \\
&=& (S_{7,3} + 4 * S_{7,4}) + 5 * (S_{7,4} + 5 * S_{7,5}) \\
&=& ((S_{6,2} + 3 * S_{6.3}) + 4 * (S_{6,3} + 4 * S_{6,4})) + 5 * ((S_{6,3} + 4 * S_{6,4}) + 5 *
(S_{6,4} + 5 * S_{6,5})) \\
&=& ((32 + 3 * 90) + 4*(90 + 4*(S_{5,3} + 4*S_{5,4}))) + 5 * ((90 + 4*(S_{5,3} + 4*S_{5,4})) \\
&&+ 5*((S_{5,3}+ 4*S_{5,4}) + 5*(S_{5,4} + 5*S_{5,5}))) \\
&=& (302 + 4*(90+4*(57 + 4*(S_{4,3} + 4*S_{4,4})))) \\
&&+ 5*((90+4*(57 + 4*(S_{4,3} + 4*S_{4,4}))) \\
&&+ 5*((57 + 4*(S_{4,3} + 4*S_{4,4})) + 5 * ((S_{4,3} + 4*S_{4,4}) + 5 * 1))) \\
&=& (302 + 4*(90 + 4*(57 + 4*(22 + 4*1)))) \\
&& + 5*((90 + 4*(57 + 4*(22 + 4*1)))) \\
&& + 5 * ((57 + 4*(22 + 4*1)) + 5*((22 + 4*1) + 5)) \\
&=& 3238 + 3670 + 1580 \\
&=& 8488
\end{eqnarray*}
\item Bälle weiterhin unterscheidbar, Urnen jetzt auch unterscheidbar $\Rightarrow$ geordnete Mengenpartition.
\begin{eqnarray*}
k! * S_{n.k} &=& 5! * S_{9,5} \\
&=& 120 * 8488 \\
&=& 1018560
\end{eqnarray*}
\item Jetzt gilt Teller $\widehat{=}$ Ball, Keks $\widehat{=}$ Urne. $n = 5$, $k = 3$.
Urnen sind unterscheidbar, \enquote{fünfgangiges Menü} $\Rightarrow$ Bälle sind auch untescheidbar
\end{enumerate}
\end{document}