866 lines
24 KiB
TeX
866 lines
24 KiB
TeX
\chapter{Combinatorics}
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\index{combinatorics}
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\key{Combinatorics} studies methods for counting
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combinations of objects.
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Usually, the goal is to find a way to
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count the combinations efficiently
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without generating each combination separately.
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As an example, let us consider the problem
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of counting the number of ways to
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represent an integer $n$ as a sum of positive integers.
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For example, there are 8 representations
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for the number $4$:
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\begin{multicols}{2}
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\begin{itemize}
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\item $1+1+1+1$
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\item $1+1+2$
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\item $1+2+1$
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\item $2+1+1$
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\item $2+2$
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\item $3+1$
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\item $1+3$
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\item $4$
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\end{itemize}
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\end{multicols}
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A combinatorial problem can often be solved
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using a recursive function.
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In this problem, we can define a function $f(n)$
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that gives the number of representations for $n$.
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For example, $f(4)=8$ according to the above example.
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The values of the function
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can be recursively calculated as follows:
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\begin{equation*}
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f(n) = \begin{cases}
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1 & n = 1\\
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f(1)+f(2)+\ldots+f(n-1)+1 & n > 1\\
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\end{cases}
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\end{equation*}
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The base case is $f(1)=1$,
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because there is only one way to represent the number 1.
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When $n>1$, we go through all ways to
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choose the last number in the sum.
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For example, in when $n=4$, the sum can end
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with $+1$, $+2$ or $+3$.
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In addition, we also count the representation
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that only contains $n$.
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The first values for the function are:
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\[
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\begin{array}{lcl}
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f(1) & = & 1 \\
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f(2) & = & 2 \\
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f(3) & = & 4 \\
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f(4) & = & 8 \\
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f(5) & = & 16 \\
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\end{array}
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\]
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It turns out that the function also has a closed-form formula
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\[
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f(n)=2^{n-1},
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\]
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which is based on the fact that there are $n-1$
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possible positions for +-signs in the sum
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and we can choose any subset of them.
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\section{Binomial coefficients}
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\index{binomial coefficient}
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The \key{binomial coefficient} ${n \choose k}$
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equals the number of ways we can choose a subset
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of $k$ elements from a set of $n$ elements.
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For example, ${5 \choose 3}=10$,
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because the set $\{1,2,3,4,5\}$
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has 10 subsets of 3 elements:
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\[ \{1,2,3\}, \{1,2,4\}, \{1,2,5\}, \{1,3,4\}, \{1,3,5\},
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\{1,4,5\}, \{2,3,4\}, \{2,3,5\}, \{2,4,5\}, \{3,4,5\} \]
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\subsubsection{Formula 1}
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Binomial coefficients can be
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recursively calculated as follows:
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\[
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{n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}
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\]
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The idea is to fix an element $x$ in the set.
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If $x$ is included in the subset,
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we have to choose $k-1$
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elements from $n-1$ elements,
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and if $x$ is not included in the subset,
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we have to choose $k$ elements from $n-1$ elements.
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The base cases for the recursion are
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\[
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{n \choose 0} = {n \choose n} = 1,
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\]
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because there is always exactly
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one way to construct an empty subset
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and a subset that contains all the elements.
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\subsubsection{Formula 2}
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Another way to calculate binomial coefficients is as follows:
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\[
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{n \choose k} = \frac{n!}{k!(n-k)!}.
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\]
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There are $n!$ permutations of $n$ elements.
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We go through all permutations and always
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select the first $k$ elements of the permutation
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to the subset.
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Since the order of the elements in the subset
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and outside the subset does not matter,
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the result is divided by $k!$ and $(n-k)!$
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\subsubsection{Properties}
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For binomial coefficients,
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\[
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{n \choose k} = {n \choose n-k},
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\]
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because we can either select $k$
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elements that belong to the subset
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or $n-k$ elements that
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do not belong to the subset.
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The sum of binomial coefficients is
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\[
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{n \choose 0}+{n \choose 1}+{n \choose 2}+\ldots+{n \choose n}=2^n.
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\]
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The reason for the name ''binomial coefficient''
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is that
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\[ (a+b)^n =
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{n \choose 0} a^n b^0 +
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{n \choose 1} a^{n-1} b^1 +
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\ldots +
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{n \choose n-1} a^1 b^{n-1} +
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{n \choose n} a^0 b^n. \]
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\index{Pascal's triangle}
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Binomial coefficients also appear in
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\key{Pascal's triangle}
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where each value equals the sum of two
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above values:
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\begin{center}
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\begin{tikzpicture}{0.9}
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\node at (0,0) {1};
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\node at (-0.5,-0.5) {1};
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\node at (0.5,-0.5) {1};
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\node at (-1,-1) {1};
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\node at (0,-1) {2};
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\node at (1,-1) {1};
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\node at (-1.5,-1.5) {1};
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\node at (-0.5,-1.5) {3};
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\node at (0.5,-1.5) {3};
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\node at (1.5,-1.5) {1};
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\node at (-2,-2) {1};
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\node at (-1,-2) {4};
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\node at (0,-2) {6};
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\node at (1,-2) {4};
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\node at (2,-2) {1};
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\node at (-2,-2.5) {$\ldots$};
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\node at (-1,-2.5) {$\ldots$};
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\node at (0,-2.5) {$\ldots$};
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\node at (1,-2.5) {$\ldots$};
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\node at (2,-2.5) {$\ldots$};
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\end{tikzpicture}
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\end{center}
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\subsubsection{Boxes and balls}
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''Boxes and balls'' is a useful model,
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where we count the ways to
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place $k$ balls in $n$ boxes.
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Let us consider three scenarios:
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\textit{Scenario 1}: Each box can contain
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at most one ball.
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For example, when $n=5$ and $k=2$,
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there are 10 solutions:
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\newcommand\lax[3]{
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\path[draw,thick,-] (#1-0.5,#2+0.5) -- (#1-0.5,#2-0.5) --
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(#1+0.5,#2-0.5) -- (#1+0.5,#2+0.5);
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\ifthenelse{\equal{#3}{1}}{\draw[fill=black] (#1,#2-0.3) circle (0.15);}{}
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\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1-0.2,#2-0.3) circle (0.15);}{}
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\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1+0.2,#2-0.3) circle (0.15);}{}
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}
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\newcommand\laa[7]{
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\lax{#1}{#2}{#3}
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\lax{#1+1.2}{#2}{#4}
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\lax{#1+2.4}{#2}{#5}
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\lax{#1+3.6}{#2}{#6}
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\lax{#1+4.8}{#2}{#7}
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}
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\laa{0}{0}{1}{1}{0}{0}{0}
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\laa{0}{-2}{1}{0}{1}{0}{0}
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\laa{0}{-4}{1}{0}{0}{1}{0}
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\laa{0}{-6}{1}{0}{0}{0}{1}
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\laa{8}{0}{0}{1}{1}{0}{0}
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\laa{8}{-2}{0}{1}{0}{1}{0}
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\laa{8}{-4}{0}{1}{0}{0}{1}
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\laa{16}{0}{0}{0}{1}{1}{0}
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\laa{16}{-2}{0}{0}{1}{0}{1}
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\laa{16}{-4}{0}{0}{0}{1}{1}
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\end{tikzpicture}
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\end{center}
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In this scenario, the answer is directly the
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binomial coefficient ${n \choose k}$.
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\textit{Scenario 2}: A box can contain multiple balls.
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For example, when $n=5$ and $k=2$,
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there are 15 solutions:
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\newcommand\lax[3]{
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\path[draw,thick,-] (#1-0.5,#2+0.5) -- (#1-0.5,#2-0.5) --
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(#1+0.5,#2-0.5) -- (#1+0.5,#2+0.5);
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\ifthenelse{\equal{#3}{1}}{\draw[fill=black] (#1,#2-0.3) circle (0.15);}{}
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\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1-0.2,#2-0.3) circle (0.15);}{}
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\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1+0.2,#2-0.3) circle (0.15);}{}
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}
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\newcommand\laa[7]{
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\lax{#1}{#2}{#3}
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\lax{#1+1.2}{#2}{#4}
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\lax{#1+2.4}{#2}{#5}
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\lax{#1+3.6}{#2}{#6}
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\lax{#1+4.8}{#2}{#7}
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}
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\laa{0}{0}{2}{0}{0}{0}{0}
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\laa{0}{-2}{1}{1}{0}{0}{0}
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\laa{0}{-4}{1}{0}{1}{0}{0}
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\laa{0}{-6}{1}{0}{0}{1}{0}
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\laa{0}{-8}{1}{0}{0}{0}{1}
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\laa{8}{0}{0}{2}{0}{0}{0}
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\laa{8}{-2}{0}{1}{1}{0}{0}
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\laa{8}{-4}{0}{1}{0}{1}{0}
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\laa{8}{-6}{0}{1}{0}{0}{1}
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\laa{8}{-8}{0}{0}{2}{0}{0}
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\laa{16}{0}{0}{0}{1}{1}{0}
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\laa{16}{-2}{0}{0}{1}{0}{1}
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\laa{16}{-4}{0}{0}{0}{2}{0}
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\laa{16}{-6}{0}{0}{0}{1}{1}
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\laa{16}{-8}{0}{0}{0}{0}{2}
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\end{tikzpicture}
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\end{center}
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The process of placing the balls in the boxes
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can be represented as a string
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that consists of symbols
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''o'' and ''$\rightarrow$''.
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Initially, assume that we are standing at the leftmost box.
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The symbol ''o'' means that we place a ball
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in the current box, and the symbol
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''$\rightarrow$'' means that we move to
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the next box to the right.
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Using this notation, each solution is a string
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that contains $k$ times the symbol ''o'' and
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$n-1$ times the symbol ''$\rightarrow$''.
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For example, the upper-right solution
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in the above picture corresponds to the string
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''$\rightarrow$ $\rightarrow$ o $\rightarrow$ o $\rightarrow$''.
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Thus, the number of solutions is
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${k+n-1 \choose k}$.
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\textit{Scenario 3}: Each box may contain at most one ball,
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and in addition, no two adjacent boxes may both contain a ball.
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For example, when $n=5$ and $k=2$,
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there are 6 solutions:
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\newcommand\lax[3]{
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\path[draw,thick,-] (#1-0.5,#2+0.5) -- (#1-0.5,#2-0.5) --
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(#1+0.5,#2-0.5) -- (#1+0.5,#2+0.5);
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\ifthenelse{\equal{#3}{1}}{\draw[fill=black] (#1,#2-0.3) circle (0.15);}{}
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\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1-0.2,#2-0.3) circle (0.15);}{}
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\ifthenelse{\equal{#3}{2}}{\draw[fill=black] (#1+0.2,#2-0.3) circle (0.15);}{}
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}
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\newcommand\laa[7]{
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\lax{#1}{#2}{#3}
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\lax{#1+1.2}{#2}{#4}
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\lax{#1+2.4}{#2}{#5}
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\lax{#1+3.6}{#2}{#6}
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\lax{#1+4.8}{#2}{#7}
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}
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\laa{0}{0}{1}{0}{1}{0}{0}
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\laa{0}{-2}{1}{0}{0}{1}{0}
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\laa{8}{0}{1}{0}{0}{0}{1}
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\laa{8}{-2}{0}{1}{0}{1}{0}
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\laa{16}{0}{0}{1}{0}{0}{1}
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\laa{16}{-2}{0}{0}{1}{0}{1}
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\end{tikzpicture}
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\end{center}
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In this scenario, we can assume that
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$k$ balls are initially placed in boxes
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and there is an empty box between each
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two such boxes.
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The remaining task is to choose the
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positions for
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$n-k-(k-1)=n-2k+1$ empty boxes.
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There are $k+1$ positions,
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so the number of solutions is
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${n-2k+1+k+1-1 \choose n-2k+1} = {n-k+1 \choose n-2k+1}$.
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\subsubsection{Multinomial coefficient}
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\index{multinomial coefficient}
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The \key{multinomial coefficient}
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\[ {n \choose k_1,k_2,\ldots,k_m} = \frac{n!}{k_1! k_2! \cdots k_m!}, \]
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equals the number of ways
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we can divide $n$ elements into subsets
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of sizes $k_1,k_2,\ldots,k_m$,
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where $k_1+k_2+\cdots+k_m=n$.
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Multinomial coefficients can be seen as a
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generalization of binomial cofficients;
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if $m=2$, the above formula
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corresponds to the binomial coefficient formula.
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\section{Catalan numbers}
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\index{Catalan number}
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The \key{Catalan number} $C_n$ equals the
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number of valid
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parenthesis expressions that consist of
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$n$ left parentheses and $n$ right parentheses.
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For example, $C_3=5$, because using three
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left parentheses and three right parentheses,
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we can construct the following parenthesis
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expressions:
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\begin{itemize}[noitemsep]
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\item \texttt{()()()}
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\item \texttt{(())()}
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\item \texttt{()(())}
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\item \texttt{((()))}
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\item \texttt{(()())}
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\end{itemize}
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\subsubsection{Parenthesis expressions}
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\index{parenthesis expression}
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What is exactly a \emph{valid parenthesis expression}?
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The following rules precisely define all
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valid parenthesis expressions:
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\begin{itemize}
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\item The empty expression is valid.
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\item If an expression $A$ is valid,
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then also the expression
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\texttt{(}$A$\texttt{)} is valid.
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\item If expressions $A$ and $B$ are valid,
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then also the expression $AB$ is valid.
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\end{itemize}
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Another way to characterize valid
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parenthesis expressions is that if
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we choose any prefix of such an expression,
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it has to contain at least as many left
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parentheses as right parentheses.
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In addition, the complete expression has to
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contain an equal number of left and right
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parentheses.
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\subsubsection{Formula 1}
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Catalan numbers can be calculated using the formula
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\[ C_n = \sum_{i=0}^{n-1} C_{i} C_{n-i-1}.\]
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The sum goes through the ways to divide the
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expression into two parts
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such that both parts are valid
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expressions and the first part is as short as possible
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but not empty.
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For any $i$, the first part contains $i+1$ pairs
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of parentheses and the number of expressions
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is the product of the following values:
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\begin{itemize}
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\item $C_{i}$: number of ways to construct an expression
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using the parentheses in the first part,
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not counting the outermost parentheses
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\item $C_{n-i-1}$: number of ways to construct an
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expression using the parentheses in the second part
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\end{itemize}
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In addition, the base case is $C_0=1$,
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because we can construct an empty parenthesis
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expression using zero pairs of parentheses.
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\subsubsection{Formula 2}
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Catalan numbers can also be calculated
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using binomial coefficients:
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\[ C_n = \frac{1}{n+1} {2n \choose n}\]
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The formula can be explained as follows:
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There are a total of ${2n \choose n}$ ways
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to construct a (not necessarily valid)
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parenthesis expression that contains $n$ left
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parentheses and $n$ right parentheses.
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Let us calculate the number of such
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expressions that are \emph{not} valid.
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If a parenthesis expression is not valid,
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it has to contain a prefix where the
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number of right parentheses exceeds the
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number of left parentheses.
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The idea is to reverse each parenthesis
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that belongs to such a prefix.
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For example, the expression
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\texttt{())()(} contains a prefix \texttt{())},
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and after reversing the prefix,
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the expression becomes \texttt{)((()(}.
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The resulting expression consists of $n+1$
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left parentheses and $n-1$ right parentheses.
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The number of such expressions is ${2n \choose n+1}$
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that equals the number of non-valid
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parenthesis expressions.
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Thus the number of valid parenthesis
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expressions can be calculated using the formula
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\[{2n \choose n}-{2n \choose n+1} = {2n \choose n} - \frac{n}{n+1} {2n \choose n} = \frac{1}{n+1} {2n \choose n}.\]
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\subsubsection{Counting trees}
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Catalan numbers are also related to trees:
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\begin{itemize}
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\item there are $C_n$ binary trees of $n$ nodes
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\item there are $C_{n-1}$ rooted trees of $n$ nodes
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\end{itemize}
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\noindent
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For example, for $C_3=5$, the binary trees are
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\path[draw,thick,-] (0,0) -- (-1,-1);
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\path[draw,thick,-] (0,0) -- (1,-1);
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\draw[fill=white] (0,0) circle (0.3);
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\draw[fill=white] (-1,-1) circle (0.3);
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\draw[fill=white] (1,-1) circle (0.3);
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\path[draw,thick,-] (4,0) -- (4-0.75,-1) -- (4-1.5,-2);
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\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{Inclusion-exclusion}
|
|
|
|
\index{inclusion-exclusion}
|
|
|
|
\key{Inclusion-exclusion} is a technique
|
|
that can be used for counting the size
|
|
of a union of sets when the sizes of
|
|
the intersections are known, and vice versa.
|
|
A simple example of the technique is the formula
|
|
\[ |A \cup B| = |A| + |B| - |A \cap B|,\]
|
|
where $A$ and $B$ are sets and $|X|$
|
|
is the size of a set $X$.
|
|
The formula can be illustrated as follows:
|
|
|
|
\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}
|
|
|
|
Our goal is to calculate
|
|
the size of the union $A \cup B$
|
|
that corresponds to the area of the region
|
|
that belongs to at least one circle.
|
|
The picture shows that we can calculate
|
|
the area of $A \cup B$ by first summing the
|
|
areas of $A$ and $B$ and then subtracting
|
|
the area of $A \cap B$.
|
|
|
|
The same idea can be applied when the number
|
|
of sets is larger.
|
|
When there are three sets, the inclusion-exclusion formula is
|
|
\[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \]
|
|
and the corresponding picture is
|
|
|
|
\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}
|
|
|
|
In the general case, the size of the
|
|
union $X_1 \cup X_2 \cup \cdots \cup X_n$
|
|
can be calculated by going through all possible
|
|
intersections that contain some of the sets $X_1,X_2,\ldots,X_n$.
|
|
If the intersection contains an odd number of sets,
|
|
its size is added to the answer,
|
|
and otherwise its size is subtracted from the answer.
|
|
|
|
Note that there are similar formulas
|
|
for calculating
|
|
the size of an intersection from the sizes of
|
|
unions. For example,
|
|
\[ |A \cap B| = |A| + |B| - |A \cup B|\]
|
|
and
|
|
\[ |A \cap B \cap C| = |A| + |B| + |C| - |A \cup B| - |A \cup C| - |B \cup C| + |A \cup B \cup C| .\]
|
|
|
|
\subsubsection{Derangements}
|
|
|
|
\index{derangement}
|
|
|
|
As an example, let us count the number of \key{derangements}
|
|
of elements $\{1,2,\ldots,n\}$, i.e., permutations
|
|
where no element remains in its original place.
|
|
For example, when $n=3$, there are
|
|
two possible derangements: $(2,3,1)$ and $(3,1,2)$.
|
|
|
|
One approach for solving the problem is to use
|
|
inclusion-exclusion.
|
|
Let $X_k$ be the set of permutations
|
|
that contain the element $k$ at position $k$.
|
|
For example, when $n=3$, the sets are as follows:
|
|
\[
|
|
\begin{array}{lcl}
|
|
X_1 & = & \{(1,2,3),(1,3,2)\} \\
|
|
X_2 & = & \{(1,2,3),(3,2,1)\} \\
|
|
X_3 & = & \{(1,2,3),(2,1,3)\} \\
|
|
\end{array}
|
|
\]
|
|
Using these sets, the number of derangements equals
|
|
\[ n! - |X_1 \cup X_2 \cup \cdots \cup X_n|, \]
|
|
so it suffices to calculate the size of the union.
|
|
Using inclusion-exclusion, this reduces to
|
|
calculating sizes of intersections which can be
|
|
done efficiently.
|
|
For example, when $n=3$, the size of
|
|
$|X_1 \cup X_2 \cup X_3|$ is
|
|
\[
|
|
\begin{array}{lcl}
|
|
& & |X_1| + |X_2| + |X_3| - |X_1 \cap X_2| - |X_1 \cap X_3| - |X_2 \cap X_3| + |X_1 \cap X_2 \cap X_3| \\
|
|
& = & 2+2+2-1-1-1+1 \\
|
|
& = & 4, \\
|
|
\end{array}
|
|
\]
|
|
so the number of solutions is $3!-4=2$.
|
|
|
|
It turns out that the problem can also be solved
|
|
without using inclusion-exclusion.
|
|
Let $f(n)$ denote the number of derangements
|
|
for $\{1,2,\ldots,n\}$. We can use the following
|
|
recursive formula:
|
|
|
|
\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*}
|
|
|
|
The formula can be derived by going through
|
|
the possibilities how the element 1 changes
|
|
in the derangement.
|
|
There are $n-1$ ways to choose an element $x$
|
|
that replaces the element 1.
|
|
In each such choice, there are two options:
|
|
|
|
\textit{Option 1:} We also replace the element $x$
|
|
with the element 1.
|
|
After this, the remaining task is to construct
|
|
a derangement of $n-2$ elements.
|
|
|
|
\textit{Option 2:} We replace the element $x$
|
|
with some other element than 1.
|
|
Now we have to construct a derangement
|
|
of $n-1$ element, because we cannot replace
|
|
the element $x$ with the element $1$, and all other
|
|
elements should be changed.
|
|
|
|
\section{Burnside's lemma}
|
|
|
|
\index{Burnside's lemma}
|
|
|
|
\key{Burnside's lemma} can be used to count
|
|
the number of combinations so that
|
|
only one representative is counted
|
|
for each group of symmetric combinations.
|
|
Burnside's lemma states that the number of
|
|
combinations is
|
|
\[\sum_{k=1}^n \frac{c(k)}{n},\]
|
|
where there are $n$ ways to change the
|
|
position of a combination,
|
|
and there are $c(k)$ combinations that
|
|
remain unchanged when the $k$th way is applied.
|
|
|
|
As an example, let us calculate the number of
|
|
necklaces of $n$ pearls,
|
|
where the color of each pearl is
|
|
one of $1,2,\ldots,m$.
|
|
Two necklaces are symmetric if they are
|
|
similar after rotating them.
|
|
For example, the necklace
|
|
\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}
|
|
has the following symmetric necklaces:
|
|
\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}
|
|
There are $n$ ways to change the position
|
|
of a necklace,
|
|
because we can rotate it
|
|
$0,1,\ldots,n-1$ steps clockwise.
|
|
If the number of steps is 0,
|
|
all $m^n$ necklaces remain the same,
|
|
and if the number of steps is 1,
|
|
only the $m$ necklaces where each
|
|
pearl has the same color remain the same.
|
|
|
|
More generally, when the number of steps is $k$,
|
|
a total of
|
|
\[m^{\textrm{gcd}(k,n)},\]
|
|
necklaces remain the same,
|
|
where $\textrm{gcd}(k,n)$ is the greatest common
|
|
divisor of $k$ and $n$.
|
|
The reason for this is that blocks
|
|
of pearls of size $\textrm{gcd}(k,n)$
|
|
will replace each other.
|
|
Thus, according to Burnside's lemma,
|
|
the number of necklaces is
|
|
\[\sum_{i=0}^{n-1} \frac{m^{\textrm{gcd}(i,n)}}{n}. \]
|
|
For example, the number of necklaces of length 4
|
|
with 3 colors is
|
|
\[\frac{3^4+3+3^2+3}{4} = 24. \]
|
|
|
|
\section{Cayley's formula}
|
|
|
|
\index{Cayley's formula}
|
|
|
|
\key{Cayley's formula} states that
|
|
there are $n^{n-2}$ labeled trees
|
|
that contain $n$ nodes.
|
|
The nodes are labeled $1,2,\ldots,n$,
|
|
and two trees are different
|
|
if either their structure or
|
|
labeling is different.
|
|
|
|
\begin{samepage}
|
|
For example, when $n=4$, the number of labeled
|
|
trees is $4^{4-2}=16$:
|
|
|
|
\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}
|
|
|
|
Next we will see how Cayley's formula can
|
|
be derived using Prüfer codes.
|
|
|
|
\subsubsection{Prüfer code}
|
|
|
|
\index{Prüfer code}
|
|
|
|
A \key{Prüfer code} is a sequence of
|
|
$n-2$ numbers that describes a labeled tree.
|
|
The code is constructed by following a process
|
|
that removes $n-2$ leaves from the tree.
|
|
At each step, the leaf with the smallest label is removed,
|
|
and the label of its only neighbor is added to the code.
|
|
|
|
For example, the Prüfer code for
|
|
\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}
|
|
is $[4,4,2]$, because we first remove
|
|
node 1, then node 3 and finally node 5.
|
|
|
|
We can construct a Prüfer code for any tree,
|
|
and more importantly,
|
|
the original tree can be reconstructed
|
|
from a Prüfer code.
|
|
Hence, the number of labeled trees
|
|
of $n$ nodes equals
|
|
$n^{n-2}$, the number of Prüfer codes
|
|
of size $n$.
|