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%%% file:      v2.tex                                                        %%%
%%% author:    Ján Komara                                                    %%%
%%% email:     komara@fmph.uniba.sk                                          %%%
%%% copyright: public domain                                                 %%%
%%% version:   2015/10/18                                                    %%%
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\begin{document}

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\title{Vzorové príklady pre 2. domácu úlohu}

\author{Ján Komara}

%\date{}
\date{\today}

\maketitle

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\section*{Príklad}

Spočítajte, koľko celočíselných riešení má rovnica
\begin{equation}
x_1 + x_2 + x_3 + x_4 = 15,
\label{eq15}
\end{equation}
ak $x_1 \geq 1$, $x_2 \geq 2$, $x_3 \geq 3$ a $x_4 \geq 4$.

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\subsection*{Riešenie}

Rovnica \eqref{eq15} je ekvivalentná rovnici
\begin{equation}
x_1 - 1 + x_2 - 2 + x_3 - 3 + x_4 - 4 = 15 - 1 - 2 - 3 - 4 = 5,
\label{eq5}
\end{equation}
pričom $x_1 - 1 \geq 0$, $x_2 - 2 \geq 0$, $x_3 - 3 \geq 0$ a $x_4 - 4 \geq 0$.

Počet celočíselných riešení rovnice \eqref{eq5} s uvedenými ohraničeniami je
ten istý ako počet celočíselných riešení rovnice
\[
y_1 + y_2 + y_3 + y_4 = 5,
\]
kde $y_i \geq 0$ pre každé $i = 1,2,3,4$.
Ale ten je rovný počtu kombinácií s opakovaním 5 prvkov zo 4 druhov:
\[
C'(4,5) =  \binom{5 + 4 - 1}{4 - 1} = \binom{8}{3} = 
\frac{8 \times 7 \times 6}{3 \times 2} = 56.
\]
A to je hľadaný počet riešení rovnice \eqref{eq15}.

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\section*{Príklad}

Určite koeficient pri $x^5 y^2 z^3$ vo výraze $(x + 4y - z)^{10}$.

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\subsection*{Riešenie}

Podľa multinomickej vety platí
\begin{align*}
(x + y + z)^{10} & =
\sum_{n_1 + n_2 + n_3 = 10} 
\binom{10}{n_1,n_2,n_3} x^{n_1} (4y)^{n_2} (-z)^{n_3}
\\ & =
\sum_{n_1 + n_2 + n_3 = 10} 
\binom{10}{n_1,n_2,n_3} 4^{n_2} (-1)^{n_3} x^{n_1} y^{n_2} z^{n_3}.
\end{align*}
Hľadaný koeficient pri $x^5 y^2 z^3$ je preto rovný číslu
\[
\binom{10}{5,2,3} 4^2 (-1)^3 = -16 \binom{10}{5,2,3}.
\]

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