\chapter{Combinatorics}

\index{combinatorics}

\key{Combinatorics} studies methods for counting
combinations of objects.
Usually, the goal is to find a way to
count the combinations efficiently
without generating each combination separately.

As an example, let's consider a problem where
our task is to calculate the number of representations
for an integer $n$ as a sum of positive integers.
For example, there are 8 representations
for the number $4$:
\begin{multicols}{2}
\begin{itemize}
\item $1+1+1+1$
\item $1+1+2$
\item $1+2+1$
\item $2+1+1$
\item $2+2$
\item $3+1$
\item $1+3$
\item $4$
\end{itemize}
\end{multicols}

A combinatorial problem can often be solved
using a recursive function.
In this case, we can define a function $f(n)$
that counts the number of representations for $n$.
For example, $f(4)=8$ according to the above example.
The function can be recursively calculated as follows:
\begin{equation*}
    f(n) = \begin{cases}
               1               & n = 1\\
               f(1)+f(2)+\ldots+f(n-1)+1 & n > 1\\
           \end{cases}
\end{equation*}
The base case is $f(1)=1$,
because there is only one way to represent the number 1.
Otherwise, we go through all possibilities for
the last number in the sum.
For example, in when $n=4$, the sum can end
with $+1$, $+2$ or $+3$.
In addition, we also count the representation
that only contains $n$.

The first values for the function are:
\[
\begin{array}{lcl}
f(1) & = & 1 \\
f(2) & = & 2 \\
f(3) & = & 4 \\
f(4) & = & 8 \\
f(5) & = & 16 \\
\end{array}
\]
It turns out that the function also has a closed-form formula
\[
f(n)=2^{n-1},
\]
which is based on the fact that there are $n-1$
possible positions for +-signs in the sum,
and we can choose any subset of them.

\section{Binomial coefficient}

\index{binomial coefficient}

A \key{binomial coefficient} ${n \choose k}$
is the number of ways we can choose a subset
of $k$ elements from a set of $n$ elements.
For example, ${5 \choose 3}=10$,
because the set $\{1,2,3,4,5\}$
has 10 subsets of 3 elements:
\[ \{1,2,3\}, \{1,2,4\}, \{1,2,5\}, \{1,3,4\}, \{1,3,5\}, 
\{1,4,5\}, \{2,3,4\}, \{2,3,5\}, \{2,4,5\}, \{3,4,5\} \]

\subsubsection{Formula 1}

Binomial coefficients can be
recursively calculated as follows:

\[
{n \choose k}  =  {n-1 \choose k-1} + {n-1 \choose k}
\]

The idea is to consider a fixed
element $x$ in the set.
If $x$ is included in the subset,
the remaining task is to choose $k-1$
elements from $n-1$ elements,
and otherwise
the remaining task is to choose $k$ elements from $n-1$ elements.

The base cases for the recursion are as follows:
\[
{n \choose 0}  =  {n \choose n} = 1
\]

The reason for this is that there is always
one way to construct an empty subset,
or a subset that contains all the elements.

\subsubsection{Formula 2}

Another way to calculate binomial coefficients is as follows:
\[
{n \choose k}  =  \frac{n!}{k!(n-k)!}.
\]

There are $n!$ permutations for $n$ elements.
We go through all permutations and in each case
select the first $k$ elements of the permutation
to the subset.
Since the order of the elements in the subset
and outside the subset doesn't matter,
the result is divided by $k!$ and $(n-k)!$

\subsubsection{Properties}

For binomial coefficients,
\[
{n \choose k}  =  {n \choose n-k},
\]
because we can either select $k$
elements to the subset,
or select $n-k$ elements that
will be outside the subset.

The sum of binomial coefficients is
\[
{n \choose 0}+{n \choose 1}+{n \choose 2}+\ldots+{n \choose n}=2^n.
\]

The reason for the name ''binomial coefficient''
is that

\[ (a+b)^n =
{n \choose 0} a^n b^0 + 
{n \choose 1} a^{n-1} b^1 +
\ldots + 
{n \choose n-1} a^1 b^{n-1} +
{n \choose n} a^0 b^n. \]

\index{Pascal's triangle}

Binomial coefficients also appear in
\key{Pascal's triangle}
whose border consists of 1's,
and each value is the sum of two
above values:
\begin{center}
\begin{tikzpicture}{0.9}
\node at (0,0) {1};
\node at (-0.5,-0.5) {1};
\node at (0.5,-0.5) {1};
\node at (-1,-1) {1};
\node at (0,-1) {2};
\node at (1,-1) {1};
\node at (-1.5,-1.5) {1};
\node at (-0.5,-1.5) {3};
\node at (0.5,-1.5) {3};
\node at (1.5,-1.5) {1};
\node at (-2,-2) {1};
\node at (-1,-2) {4};
\node at (0,-2) {6};
\node at (1,-2) {4};
\node at (2,-2) {1};
\node at (-2,-2.5) {$\ldots$};
\node at (-1,-2.5) {$\ldots$};
\node at (0,-2.5) {$\ldots$};
\node at (1,-2.5) {$\ldots$};
\node at (2,-2.5) {$\ldots$};
\end{tikzpicture}
\end{center}

\subsubsection{Boxes and balls}

''Boxes and models'' is a useful model,
where we count the ways to
place $k$ balls in $n$ boxes.
Let's consider three cases:

\textit{Case 1}: Each box can contain
at most one ball.
For example, when $n=5$ and $k=2$,
there are 10 solutions:

\begin{center}
\begin{tikzpicture}[scale=0.5]
\newcommand\lax[3]{
\path[draw,thick,-] (#1-0.5,#2+0.5) -- (#1-0.5,#2-0.5) --
                    (#1+0.5,#2-0.5) -- (#1+0.5,#2+0.5);
\ifthenelse{\equal{#3}{1}}{\draw[fill=black] (#1,#2-0.3) circle (0.15);}{}
\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1-0.2,#2-0.3) circle (0.15);}{}
\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1+0.2,#2-0.3) circle (0.15);}{}
}
\newcommand\laa[7]{
    \lax{#1}{#2}{#3}
    \lax{#1+1.2}{#2}{#4}
    \lax{#1+2.4}{#2}{#5}
    \lax{#1+3.6}{#2}{#6}
    \lax{#1+4.8}{#2}{#7}
}

\laa{0}{0}{1}{1}{0}{0}{0}
\laa{0}{-2}{1}{0}{1}{0}{0}
\laa{0}{-4}{1}{0}{0}{1}{0}
\laa{0}{-6}{1}{0}{0}{0}{1}
\laa{8}{0}{0}{1}{1}{0}{0}
\laa{8}{-2}{0}{1}{0}{1}{0}
\laa{8}{-4}{0}{1}{0}{0}{1}
\laa{16}{0}{0}{0}{1}{1}{0}
\laa{16}{-2}{0}{0}{1}{0}{1}
\laa{16}{-4}{0}{0}{0}{1}{1}

\end{tikzpicture}
\end{center}

In this case, the answer is directly the
binomial coefficient ${n \choose k}$.

\textit{Case 2}: A box can contain multiple balls.
For example, when $n=5$ and $k=2$,
there are 15 solutions:

\begin{center}
\begin{tikzpicture}[scale=0.5]
\newcommand\lax[3]{
\path[draw,thick,-] (#1-0.5,#2+0.5) -- (#1-0.5,#2-0.5) --
                    (#1+0.5,#2-0.5) -- (#1+0.5,#2+0.5);
\ifthenelse{\equal{#3}{1}}{\draw[fill=black] (#1,#2-0.3) circle (0.15);}{}
\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1-0.2,#2-0.3) circle (0.15);}{}
\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1+0.2,#2-0.3) circle (0.15);}{}
}
\newcommand\laa[7]{
    \lax{#1}{#2}{#3}
    \lax{#1+1.2}{#2}{#4}
    \lax{#1+2.4}{#2}{#5}
    \lax{#1+3.6}{#2}{#6}
    \lax{#1+4.8}{#2}{#7}
}

\laa{0}{0}{2}{0}{0}{0}{0}
\laa{0}{-2}{1}{1}{0}{0}{0}
\laa{0}{-4}{1}{0}{1}{0}{0}
\laa{0}{-6}{1}{0}{0}{1}{0}
\laa{0}{-8}{1}{0}{0}{0}{1}
\laa{8}{0}{0}{2}{0}{0}{0}
\laa{8}{-2}{0}{1}{1}{0}{0}
\laa{8}{-4}{0}{1}{0}{1}{0}
\laa{8}{-6}{0}{1}{0}{0}{1}
\laa{8}{-8}{0}{0}{2}{0}{0}
\laa{16}{0}{0}{0}{1}{1}{0}
\laa{16}{-2}{0}{0}{1}{0}{1}
\laa{16}{-4}{0}{0}{0}{2}{0}
\laa{16}{-6}{0}{0}{0}{1}{1}
\laa{16}{-8}{0}{0}{0}{0}{2}

\end{tikzpicture}
\end{center}

This process can be represented as a string
that consists of symbols
''o'' and ''$\rightarrow$''.
Initially, we are standing at the leftmost box.
The symbol ''o'' means we place a ball
in the current box, and the symbol
''$\rightarrow$'' means that we move to
the next box right.

Using this notation, each solution is a string
that has $k$ times the symbol ''o'' and
$n-1$ times the symbol ''$\rightarrow$''.
For example, the upper-right solution
corresponds to the string
''$\rightarrow$ $\rightarrow$ o $\rightarrow$ o $\rightarrow$''.
Thus, the number of solutions is
${k+n-1 \choose k}$.

\textit{Case 3}: Each box may contain at most one ball,
and in addition, no two adjacent boxes may both contain a ball.
For example, when $n=5$ and $k=2$,
there are 6 solutions:


\begin{center}
\begin{tikzpicture}[scale=0.5]
\newcommand\lax[3]{
\path[draw,thick,-] (#1-0.5,#2+0.5) -- (#1-0.5,#2-0.5) --
                    (#1+0.5,#2-0.5) -- (#1+0.5,#2+0.5);
\ifthenelse{\equal{#3}{1}}{\draw[fill=black] (#1,#2-0.3) circle (0.15);}{}
\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1-0.2,#2-0.3) circle (0.15);}{}
\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1+0.2,#2-0.3) circle (0.15);}{}
}
\newcommand\laa[7]{
    \lax{#1}{#2}{#3}
    \lax{#1+1.2}{#2}{#4}
    \lax{#1+2.4}{#2}{#5}
    \lax{#1+3.6}{#2}{#6}
    \lax{#1+4.8}{#2}{#7}
}

\laa{0}{0}{1}{0}{1}{0}{0}
\laa{0}{-2}{1}{0}{0}{1}{0}
\laa{8}{0}{1}{0}{0}{0}{1}
\laa{8}{-2}{0}{1}{0}{1}{0}
\laa{16}{0}{0}{1}{0}{0}{1}
\laa{16}{-2}{0}{0}{1}{0}{1}
\end{tikzpicture}
\end{center}

In this case, we can think that
$k$ balls are initially placed in boxes.
and between each such box there is an empty box.
The remaining task is to choose the
positions for
$n-k-(k-1)=n-2k+1$ empty boxes.
There are $k+1$ positions, so as in case 2,
the number of solutions is
${n-2k+1+k+1-1 \choose n-2k+1} = {n-k+1 \choose n-2k+1}$.

\subsubsection{Multinomial coefficient}

\index{multinomial coefficient}

A generalization for a binomial coefficient is
a \key{multinomial coefficient}

\[ {n \choose k_1,k_2,\ldots,k_m} = \frac{n!}{k_1! k_2! \cdots k_m!}, \]

where $k_1+k_2+\cdots+k_m=n$.
A multinomial coefficient i the number of ways
we can divide $n$ elements into subsets
whose sizes are $k_1,k_2,\ldots,k_m$.
If $m=2$, the formula
corresponds to the binomial coefficient formula.

\section{Catalan numbers}

\index{Catalan number}

A \key{Catalan number} $C_n$ is the
number of valid
parenthesis expressions that consist of
$n$ left parentheses and $n$ right parentheses.

For example, $C_3=5$, because using three
left parentheses and three right parentheses,
we can construct the following parenthesis
expressions:

\begin{itemize}[noitemsep]
\item \texttt{()()()}
\item \texttt{(())()}
\item \texttt{()(())}
\item \texttt{((()))}
\item \texttt{(()())}
\end{itemize}

\subsubsection{Parenthesis expressions}

\index{parenthesis expression}

What is exactly a \emph{valid parenthesis expression}?
The following rules precisely define all
valid parenthesis expressions:

\begin{itemize}
\item The expression \texttt{()} is valid.
\item If a expression $A$ is valid,
then also the expression
\texttt{(}$A$\texttt{)} is valid.
\item If expressions $A$ and $B$ are valid,
then also the expression $AB$ is valid.
\end{itemize}

Another way to characterize valid 
paranthesis expressions is that if
we choose any prefix of the expression,
it has to contain at least as many left
parentheses as right parentheses.
In addition, the complete expression has to
contain an equal number of left and right
parentheses.

\subsubsection{Formula 1}

Catalan numbers can be calculated using the formula
\[ C_n = \sum_{i=0}^{n-1} C_{i} C_{n-i-1}.\]

The sum goes through the ways to divide the
expression into two parts
such that both parts are valid
expressions and the first part is as short as possible
but not empty.
For any $i$, the first part contains $i+1$ pairs
of parentheses, and the number of expressions
is the product of the following values:

\begin{itemize}
\item $C_{i}$: number of ways to construct an expression
using the parentheses in the first part,
not counting the outermost parentheses
\item $C_{n-i-1}$: number of ways to construct an
expression using the parentheses in the second part
\end{itemize}
In addition, the base case is $C_0=1$,
because we can construct an empty parenthesis
expression using zero pairs of parentheses.

\subsubsection{Formula 2}

Catalan numbers can also be calculated
using binomial coefficients:
\[ C_n = \frac{1}{n+1} {2n \choose n}\]
The formula can be explained as follows:

There are a total of ${2n \choose n}$ ways
to construct a (not necessarily valid)
parenthesis expression that contains $n$ left
parentheses and $n$ right parentheses.
Let's calculate the number of such
expressions that are \emph{not} valid.

If a parenthesis expression is not valid,
it has to contain a prefix where the
number of right parentheses exceeds the
number of left parentheses.
The idea is to reverse each parenthesis
that belongs to such a prefix.
For example, the expression
\texttt{())()(} contains a prefix \texttt{())},
and after reversing the prefix,
the expression becomes \texttt{)((()(}.

The resulting expression consists of $n+1$
left parentheses and $n-1$ right parentheses.
The number of such expressions is ${2n \choose n+1}$
that equals the number of non-valid
parenthesis expressions.
Thus the number of valid parenthesis
expressions can be calculated using the formula
\[{2n \choose n}-{2n \choose n+1} = {2n \choose n} - \frac{n}{n+1} {2n \choose n} = \frac{1}{n+1} {2n \choose n}.\]

\subsubsection{Counting trees}

Catalan numbers are also related to rooted trees:

\begin{itemize}
\item there are $C_n$ binary trees of $n$ nodes
\item there are $C_{n-1}$ rooted trees of $n$ nodes
\end{itemize}
\noindent
For example, for $C_3=5$, the binary trees are

\begin{center}
\begin{tikzpicture}[scale=0.7]
\path[draw,thick,-] (0,0) -- (-1,-1);
\path[draw,thick,-] (0,0) -- (1,-1);
\draw[fill=white] (0,0) circle (0.3);
\draw[fill=white] (-1,-1) circle (0.3);
\draw[fill=white] (1,-1) circle (0.3);

\path[draw,thick,-] (4,0) -- (4-0.75,-1) -- (4-1.5,-2);
\draw[fill=white] (4,0) circle (0.3);
\draw[fill=white] (4-0.75,-1) circle (0.3);
\draw[fill=white] (4-1.5,-2) circle (0.3);

\path[draw,thick,-] (6.5,0) -- (6.5-0.75,-1) -- (6.5-0,-2);
\draw[fill=white] (6.5,0) circle (0.3);
\draw[fill=white] (6.5-0.75,-1) circle (0.3);
\draw[fill=white] (6.5-0,-2) circle (0.3);

\path[draw,thick,-] (9,0) -- (9+0.75,-1) -- (9-0,-2);
\draw[fill=white] (9,0) circle (0.3);
\draw[fill=white] (9+0.75,-1) circle (0.3);
\draw[fill=white] (9-0,-2) circle (0.3);

\path[draw,thick,-] (11.5,0) -- (11.5+0.75,-1) -- (11.5+1.5,-2);
\draw[fill=white] (11.5,0) circle (0.3);
\draw[fill=white] (11.5+0.75,-1) circle (0.3);
\draw[fill=white] (11.5+1.5,-2) circle (0.3);
\end{tikzpicture}
\end{center}
and the rooted trees are
\begin{center}
\begin{tikzpicture}[scale=0.7]
\path[draw,thick,-] (0,0) -- (-1,-1);
\path[draw,thick,-] (0,0) -- (0,-1);
\path[draw,thick,-] (0,0) -- (1,-1);
\draw[fill=white] (0,0) circle (0.3);
\draw[fill=white] (-1,-1) circle (0.3);
\draw[fill=white] (0,-1) circle (0.3);
\draw[fill=white] (1,-1) circle (0.3);

\path[draw,thick,-] (3,0) -- (3,-1) -- (3,-2) -- (3,-3);
\draw[fill=white] (3,0) circle (0.3);
\draw[fill=white] (3,-1) circle (0.3);
\draw[fill=white] (3,-2) circle (0.3);
\draw[fill=white] (3,-3) circle (0.3);

\path[draw,thick,-] (6+0,0) -- (6-1,-1);
\path[draw,thick,-] (6+0,0) -- (6+1,-1) -- (6+1,-2);
\draw[fill=white] (6+0,0) circle (0.3);
\draw[fill=white] (6-1,-1) circle (0.3);
\draw[fill=white] (6+1,-1) circle (0.3);
\draw[fill=white] (6+1,-2) circle (0.3);

\path[draw,thick,-] (9+0,0) -- (9+1,-1);
\path[draw,thick,-] (9+0,0) -- (9-1,-1) -- (9-1,-2);
\draw[fill=white] (9+0,0) circle (0.3);
\draw[fill=white] (9+1,-1) circle (0.3);
\draw[fill=white] (9-1,-1) circle (0.3);
\draw[fill=white] (9-1,-2) circle (0.3);

\path[draw,thick,-] (12+0,0) -- (12+0,-1) -- (12-1,-2);
\path[draw,thick,-] (12+0,0) -- (12+0,-1) -- (12+1,-2);
\draw[fill=white] (12+0,0) circle (0.3);
\draw[fill=white] (12+0,-1) circle (0.3);
\draw[fill=white] (12-1,-2) circle (0.3);
\draw[fill=white] (12+1,-2) circle (0.3);

\end{tikzpicture}
\end{center}


\section{Inkluusio-ekskluusio}

\index{inkluusio-ekskluusio}

\key{Inkluusio-ekskluusio}
on tekniikka, jonka avulla pystyy laskemaan
joukkojen yhdisteen koon leikkausten
kokojen perusteella ja päinvastoin.
Yksinkertainen esimerkki periaatteesta on kaava
\[ |A \cup B| = |A| + |B| - |A \cap B|,\]
jossa $A$ ja $B$ ovat joukkoja ja $|X|$
tarkoittaa joukon $X$ kokoa.
Seuraava kuva havainnollistaa kaavaa,
kun joukot ovat tason ympyröitä:

\begin{center}
\begin{tikzpicture}[scale=0.8]

\draw (0,0) circle (1.5);
\draw (1.5,0) circle (1.5);

\node at (-0.75,0) {\small $A$};
\node at (2.25,0) {\small $B$};
\node at (0.75,0) {\small $A \cap B$};

\end{tikzpicture}
\end{center}

Tavoitteena on laskea, kuinka suuri on yhdiste $A \cup B$
eli alue, joka on toisen tai kummankin ympyrän sisällä.
Kuvan mukaisesti yhdisteen $A \cup B$ koko
saadaan laskemalla ensin yhteen ympyröiden $A$ ja $B$ koot
ja vähentämällä siitä sitten leikkauksen $A \cap B$ koko.

Samaa ideaa voi soveltaa, kun joukkoja on enemmän.
Kolmen joukon tapauksessa kaavasta tulee
\[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B|  - |A \cap C|  - |B \cap C| + |A \cap B \cap C| \]
ja vastaava kuva on

\begin{center}
\begin{tikzpicture}[scale=0.8]

\draw (0,0) circle (1.75);
\draw (2,0) circle (1.75);
\draw (1,1.5) circle (1.75);

\node at (-0.75,-0.25) {\small $A$};
\node at (2.75,-0.25) {\small $B$};
\node at (1,2.5) {\small $C$};
\node at (1,-0.5) {\small $A \cap B$};
\node at (0,1.25) {\small $A \cap C$};
\node at (2,1.25) {\small $B \cap C$};
\node at (1,0.5) {\scriptsize $A \cap B \cap C$};

\end{tikzpicture}
\end{center}

Yleisessä tapauksessa yhdisteen $X_1 \cup X_2 \cup \cdots \cup X_n$
koon saa laskettua käymällä läpi kaikki tavat muodostaa
leikkaus joukoista $X_1,X_2,\ldots,X_n$.
Parittoman määrän joukkoja sisältävät leikkaukset
lasketaan mukaan positiivisina ja
parillisen määrän negatiivisina.

Huomaa, että vastaavat kaavat toimivat myös käänteisesti
leikkauksen koon laskemiseen yhdisteiden kokojen perusteella.
Esimerkiksi
\[ |A \cap B| = |A| + |B| - |A \cup B|\]
ja
\[ |A \cap B \cap C| = |A| + |B| + |C| - |A \cup B|  - |A \cup C|  - |B \cup C| + |A \cup B \cup C| .\]

\subsubsection{Epäjärjestykset}

\index{epxjxrjestys@epäjärjestys}

Lasketaan esimerkkinä,
montako tapaa on muodostaa luvuista
$(1,2,\ldots,n)$ \key{epäjärjestys}
eli permutaatio,
jossa mikään luku ei ole alkuperäisellä paikallaan.
Esimerkiksi jos $n=3$, niin epäjärjestyksiä on kaksi: $(2,3,1)$ ja $(3,1,2)$.

Yksi tapa lähestyä tehtävää on käyttää inkluusio-ekskluusiota.
Olkoon joukko $X_k$ niiden permutaatioiden joukko,
jossa kohdassa $k$ on luku $k$.
Esimerkiksi jos $n=3$, niin joukot ovat seuraavat:
\[
\begin{array}{lcl}
X_1 & = & \{(1,2,3),(1,3,2)\} \\
X_2 & = & \{(1,2,3),(3,2,1)\} \\
X_3 & = & \{(1,2,3),(2,1,3)\} \\
\end{array}
\]
Näitä joukkoja käyttäen epäjärjestysten määrä on
\[ n! - |X_1 \cup X_2 \cup \cdots \cup X_n|, \]
eli
riittää laskea joukkojen yhdisteen koko.
Tämä palautuu inkluusio-eks\-kluu\-sion avulla
joukkojen leikkausten kokojen laskemiseen,
mikä onnistuu tehokkaasti.
Esimerkiksi kun $n=3$, joukon $|X_1 \cup X_2 \cup X_3|$ koko on
\[
\begin{array}{lcl}
 & & |X_1| + |X_2| + |X_3| - |X_1 \cap X_2|  - |X_1 \cap X_3|  - |X_2 \cap X_3| + |X_1 \cap X_2 \cap X_3| \\
 & = & 2+2+2-1-1-1+1 \\
 & = & 4, \\
\end{array}
\]
joten ratkaisujen määrä on $3!-4=2$.

Osoittautuu, että tehtävän voi ratkaista myös toisella
tavalla käyttämättä inkluusio-ekskluusiota.
Merkitään $f(n)$:llä jonon $(1,2,\ldots,n)$ epäjärjestysten määrää,
jolloin seuraava rekursio pätee:

\begin{equation*}
    f(n) = \begin{cases}
               0               & n = 1\\
               1               & n = 2\\
               (n-1)(f(n-2) + f(n-1)) & n>2 \\
           \end{cases}
\end{equation*}

Kaavan voi perustella käymällä läpi tapaukset,
miten luku 1 muuttuu epäjärjestyksessä.
On $n-1$ tapaa valita jokin luku $x$ luvun 1 tilalle.
Jokaisessa tällaisessa valinnassa on kaksi vaihtoehtoa:

\textit{Vaihtoehto 1:} Luvun $x$ tilalle valitaan luku 1.
Tällöin jää $n-2$ lukua, joille tulee muodostaa epäjärjestys.

\textit{Vaihtoehto 2:} Luvun $x$ tilalle ei valita lukua 1.
Tällöin jää $n-1$ lukua, joille tulee muodostaa epäjärjestys,
koska luvun $x$ tilalle ei saa valita lukua 1
ja kaikki muut luvut tulee saattaa epäjärjestykseen.

\section{Burnsiden lemma}

\index{Burnsiden lemma@Burnsiden lemma}

\key{Burnsiden lemma} laskee yhdistelmien määrän niin,
että symmetrisistä yhdistelmistä lasketaan
mukaan vain yksi edustaja.
Burnsiden lemman mukaan yhdistelmien määrä on
\[\sum_{k=1}^n \frac{c(k)}{n},\]
missä yhdistelmän asentoa voi muuttaa $n$ tavalla
ja $c(k)$ on niiden yhdistelmien määrä,
jotka pysyvät ennallaan, kun asentoa
muutetaan tavalla $k$.

Lasketaan esimerkkinä, montako
erilaista tapaa on
muodostaa $n$ helmen helminauha,
kun kunkin helmen värin tulee olla
väliltä $1,2,\ldots,m$.
Kaksi helminauhaa ovat symmetriset,
jos ne voi saada näyttämään samalta pyörittämällä.
Esimerkiksi helminauhan
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw[fill=white] (0,0) circle (1);
\draw[fill=red] (0,1) circle (0.3);
\draw[fill=blue] (1,0) circle (0.3);
\draw[fill=red] (0,-1) circle (0.3);
\draw[fill=green] (-1,0) circle (0.3);
\end{tikzpicture}
\end{center}
kanssa symmetriset helminauhat ovat seuraavat:
\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw[fill=white] (0,0) circle (1);
\draw[fill=red] (0,1) circle (0.3);
\draw[fill=blue] (1,0) circle (0.3);
\draw[fill=red] (0,-1) circle (0.3);
\draw[fill=green] (-1,0) circle (0.3);

\draw[fill=white] (4,0) circle (1);
\draw[fill=green] (4+0,1) circle (0.3);
\draw[fill=red] (4+1,0) circle (0.3);
\draw[fill=blue] (4+0,-1) circle (0.3);
\draw[fill=red] (4+-1,0) circle (0.3);

\draw[fill=white] (8,0) circle (1);
\draw[fill=red] (8+0,1) circle (0.3);
\draw[fill=green] (8+1,0) circle (0.3);
\draw[fill=red] (8+0,-1) circle (0.3);
\draw[fill=blue] (8+-1,0) circle (0.3);

\draw[fill=white] (12,0) circle (1);
\draw[fill=blue] (12+0,1) circle (0.3);
\draw[fill=red] (12+1,0) circle (0.3);
\draw[fill=green] (12+0,-1) circle (0.3);
\draw[fill=red] (12+-1,0) circle (0.3);
\end{tikzpicture}
\end{center}
Tapoja muuttaa asentoa on $n$,
koska helminauhaa voi pyörittää $0,1,\ldots,n-1$
askelta myötäpäivään.
Jos helminauhaa pyörittää 0 askelta,
kaikki $m^n$ väritystä säilyvät ennallaan.
Jos taas helminauhaa pyörittää 1 askeleen,
vain $m$ yksiväristä helminauhaa säilyy ennallaan.

Yleisemmin kun helminauhaa pyörittää $k$ askelta,
ennallaan säilyvien yhdistelmien määrä on
\[m^{\textrm{syt}(k,n)},\]
missä $\textrm{syt}(k,n)$ on lukujen $k$ ja $n$
suurin yhteinen tekijä.
Tämä johtuu siitä, että $\textrm{syt}(k,n)$-kokoiset
pätkät helmiä siirtyvät toistensa paikoille
$k$ askelta eteenpäin.
Niinpä helminauhojen määrä on
Burnsiden lemman mukaan
\[\sum_{i=0}^{n-1} \frac{m^{\textrm{syt}(i,n)}}{n}. \]
Esimerkiksi kun helminauhan pituus on 4
ja värejä on 3, helminauhoja on
\[\frac{3^4+3+3^2+3}{4} = 24. \]

\section{Cayleyn kaava}

\index{Cayleyn kaava@Cayleyn kaava}

\key{Cayleyn kaavan} mukaan $n$ solmusta voi
muodostaa $n^{n-2}$ numeroitua puuta.
Puun solmut on numeroitu $1,2,\ldots,n$,
ja kaksi puuta ovat erilaiset,
jos niiden rakenne on erilainen
tai niissä on eri numerointi.

\begin{samepage}
\noindent
Esimerkiksi kun $n=4$, numeroitujen puiden määrä on $4^{4-2}=16$:

\begin{center}
\begin{tikzpicture}[scale=0.8]
\footnotesize

\newcommand\puua[6]{
\path[draw,thick,-] (#1,#2) -- (#1-1.25,#2-1.5);
\path[draw,thick,-] (#1,#2) -- (#1,#2-1.5);
\path[draw,thick,-] (#1,#2) -- (#1+1.25,#2-1.5);
\node[draw, circle, fill=white] at (#1,#2) {#3};
\node[draw, circle, fill=white] at (#1-1.25,#2-1.5) {#4};
\node[draw, circle, fill=white] at (#1,#2-1.5) {#5};
\node[draw, circle, fill=white] at (#1+1.25,#2-1.5) {#6};
}
\newcommand\puub[6]{
\path[draw,thick,-] (#1,#2) -- (#1+1,#2);
\path[draw,thick,-] (#1+1,#2) -- (#1+2,#2);
\path[draw,thick,-] (#1+2,#2) -- (#1+3,#2);
\node[draw, circle, fill=white] at (#1,#2) {#3};
\node[draw, circle, fill=white] at (#1+1,#2) {#4};
\node[draw, circle, fill=white] at (#1+2,#2) {#5};
\node[draw, circle, fill=white] at (#1+3,#2) {#6};
}

\puua{0}{0}{1}{2}{3}{4}
\puua{4}{0}{2}{1}{3}{4}
\puua{8}{0}{3}{1}{2}{4}
\puua{12}{0}{4}{1}{2}{3}

\puub{0}{-3}{1}{2}{3}{4}
\puub{4.5}{-3}{1}{2}{4}{3}
\puub{9}{-3}{1}{3}{2}{4}
\puub{0}{-4.5}{1}{3}{4}{2}
\puub{4.5}{-4.5}{1}{4}{2}{3}
\puub{9}{-4.5}{1}{4}{3}{2}
\puub{0}{-6}{2}{1}{3}{4}
\puub{4.5}{-6}{2}{1}{4}{3}
\puub{9}{-6}{2}{3}{1}{4}
\puub{0}{-7.5}{2}{4}{1}{3}
\puub{4.5}{-7.5}{3}{1}{2}{4}
\puub{9}{-7.5}{3}{2}{1}{4}
\end{tikzpicture}
\end{center}
\end{samepage}

Seuraavaksi näemme, miten Cayleyn kaavan
voi perustella samastamalla numeroidut puut
Prüfer-koodeihin.

\subsubsection{Prüfer-koodi}

\index{Prüfer-koodi}

\key{Prüfer-koodi} on $n-2$ luvun jono,
joka kuvaa numeroidun puun rakenteen.
Koodi muodostuu poistamalla puusta
joka askeleella lehden, jonka numero on pienin,
ja lisäämällä lehden vieressä olevan solmun
numeron koodiin.

Esimerkiksi puun
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (2,3) {$1$};
\node[draw, circle] (2) at (4,3) {$2$};
\node[draw, circle] (3) at (2,1) {$3$};
\node[draw, circle] (4) at (4,1) {$4$};
\node[draw, circle] (5) at (5.5,2) {$5$};

%\path[draw,thick,-] (1) -- (2);
%\path[draw,thick,-] (1) -- (3);
\path[draw,thick,-] (1) -- (4);
\path[draw,thick,-] (3) -- (4);
\path[draw,thick,-] (2) -- (4);
\path[draw,thick,-] (2) -- (5);
%\path[draw,thick,-] (4) -- (5);
\end{tikzpicture}
\end{center}
Prüfer-koodi on $[4,4,2]$,
koska puusta poistetaan ensin solmu 1,
sitten solmu 3 ja lopuksi solmu 5.

Jokaiselle puulle voidaan laskea
Prüfer-koodi, minkä lisäksi
Prüfer-koodista pystyy palauttamaan
yksikäsitteisesti alkuperäisen puun.
Niinpä numeroituja puita on yhtä monta
kuin Prüfer-koodeja eli $n^{n-2}$.