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Antti H S Laaksonen
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luku22.tex
212
luku22.tex
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@ -8,9 +8,9 @@ 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's consider a problem where
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our task is to calculate the number of representations
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for an integer $n$ as a sum of positive integers.
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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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@ -28,10 +28,11 @@ for the number $4$:
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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 case, we can define a function $f(n)$
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In this problem, we can define a function $f(n)$
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that counts 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 function can be recursively calculated as follows:
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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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@ -40,8 +41,8 @@ The function can be recursively calculated as follows:
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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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Otherwise, we go through all possibilities for
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the last number in the sum.
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When $n>1$, we go through all ways to
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select 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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@ -69,8 +70,8 @@ and we can choose any subset of them.
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\index{binomial coefficient}
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A \key{binomial coefficient} ${n \choose k}$
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is the number of ways we can choose a subset
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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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@ -87,22 +88,20 @@ recursively calculated as follows:
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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 consider a fixed
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element $x$ in the set.
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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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the remaining task is to choose $k-1$
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we have to choose $k-1$
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elements from $n-1$ elements,
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and otherwise
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the remaining task is to choose $k$ 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 as follows:
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The base cases for the recursion are
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\[
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{n \choose 0} = {n \choose n} = 1
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{n \choose 0} = {n \choose n} = 1,
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\]
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The reason for this is that there is always
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one way to construct an empty subset,
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or a subset that contains all the elements.
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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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@ -111,12 +110,12 @@ Another way to calculate binomial coefficients is as follows:
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{n \choose k} = \frac{n!}{k!(n-k)!}.
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\]
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There are $n!$ permutations for $n$ elements.
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We go through all permutations and in each case
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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 doesn't matter,
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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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@ -126,9 +125,9 @@ For binomial coefficients,
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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 to the subset,
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or select $n-k$ elements that
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will be outside the subset.
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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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@ -149,8 +148,7 @@ is that
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Binomial coefficients also appear in
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\key{Pascal's triangle}
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whose border consists of 1's,
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and each value is the sum of two
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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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@ -179,12 +177,12 @@ above values:
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\subsubsection{Boxes and balls}
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''Boxes and models'' is a useful model,
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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's consider three cases:
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Let us consider three scenarios:
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\textit{Case 1}: Each box can contain
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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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@ -220,10 +218,10 @@ there are 10 solutions:
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\end{tikzpicture}
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\end{center}
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In this case, the answer is directly the
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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{Case 2}: A box can contain multiple balls.
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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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@ -263,25 +261,26 @@ there are 15 solutions:
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\end{tikzpicture}
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\end{center}
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This process can be represented as a string
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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, we are standing at the leftmost box.
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The symbol ''o'' means we place a ball
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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 right.
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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 has $k$ times the symbol ''o'' and
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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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corresponds to the string
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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{Case 3}: Each box may contain at most one ball,
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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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@ -313,37 +312,37 @@ there are 6 solutions:
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\end{tikzpicture}
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\end{center}
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In this case, we can think that
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$k$ balls are initially placed in boxes.
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and between each such box there is an empty box.
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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, so as in case 2,
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the number of solutions is
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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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A generalization for a binomial coefficient is
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a \key{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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where $k_1+k_2+\cdots+k_m=n$.
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A multinomial coefficient i the number of ways
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equals the number of ways
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we can divide $n$ elements into subsets
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whose sizes are $k_1,k_2,\ldots,k_m$.
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If $m=2$, the formula
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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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A \key{Catalan number} $C_n$ is the
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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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@ -379,8 +378,8 @@ 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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paranthesis expressions is that if
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we choose any prefix of the expression,
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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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@ -423,7 +422,7 @@ 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's calculate the number of such
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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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@ -448,7 +447,7 @@ expressions can be calculated using the formula
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\subsubsection{Counting trees}
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Catalan numbers are also related to rooted 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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@ -554,18 +553,18 @@ The formula can be illustrated as follows:
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\end{tikzpicture}
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\end{center}
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In the above example, our goal is to calculate
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Our goal is to calculate
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the size of the union $A \cup B$
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that corresponds to the area of the region
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that is inside at least one circle.
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that belongs to at least one circle.
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The picture shows that we can calculate
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the area of $A \cup B$ by first summing the
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areas of $A$ and $B$, and then subtracting
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the area of $A \cap B$.
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The same idea can be applied, when the number
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The same idea can be applied when the number
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of sets is larger.
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When there are three sets, the formula becomes
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When there are three sets, the inclusio-exclusion formula is
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\[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \]
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and the corresponding picture is
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@ -589,14 +588,14 @@ and the corresponding picture is
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In the general case, the size of the
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union $X_1 \cup X_2 \cup \cdots \cup X_n$
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can be calculated by going through all ways to
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construct an intersection for a collection of
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sets $X_1,X_2,\ldots,X_n$.
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can be calculated by going through all possible
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intersections that contain some of the sets $X_1,X_2,\ldots,X_n$.
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If the intersection contains an odd number of sets,
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its size will be added to the answer,
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and otherwise subtracted from the answer.
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its size is added to the answer,
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and otherwise its size is subtracted from the answer.
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Note that similar formulas also work when counting
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Note that similar formulas can also be used
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for calculating
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the size of an intersection from the sizes of
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unions. For example,
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\[ |A \cap B| = |A| + |B| - |A \cup B|\]
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@ -607,16 +606,16 @@ and
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\index{derangement}
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As an example, let's count the number of \key{derangements}
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of numbers $\{1,2,\ldots,n\}$, i.e., permutations
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As an example, let us count the number of \key{derangements}
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of elements $\{1,2,\ldots,n\}$, i.e., permutations
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where no element remains in its original place.
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For example, when $n=3$, there are
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two possible derangements: $(2,3,1)$ ja $(3,1,2)$.
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two possible derangements: $(2,3,1)$ and $(3,1,2)$.
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One approach for the problem is to use
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One approach for solving the problem is to use
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inclusion-exclusion.
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Let $X_k$ be the set of permutations
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that contain the number $k$ at index $k$.
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that contain the element $k$ at position $k$.
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For example, when $n=3$, the sets are as follows:
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\[
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\begin{array}{lcl}
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@ -625,7 +624,7 @@ X_2 & = & \{(1,2,3),(3,2,1)\} \\
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X_3 & = & \{(1,2,3),(2,1,3)\} \\
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\end{array}
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\]
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Using these sets the number of derangements is
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Using these sets, the number of derangements equals
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\[ n! - |X_1 \cup X_2 \cup \cdots \cup X_n|, \]
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so it suffices to calculate the size of the union.
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Using inclusion-exclusion, this reduces to
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@ -642,8 +641,8 @@ $|X_1 \cup X_2 \cup X_3|$ is
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\]
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so the number of solutions is $3!-4=2$.
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It turns out that there is also another way for
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solving the problem without inclusion-exclusion.
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It turns out that the problem can also be solved
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without using inclusion-exclusion.
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Let $f(n)$ denote the number of derangements
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for $\{1,2,\ldots,n\}$. We can use the following
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recursive formula:
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@ -657,31 +656,32 @@ recursive formula:
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\end{equation*}
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The formula can be derived by going through
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the possibilities how the number 1 changes
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the possibilities how the element 1 changes
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in the derangement.
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There are $n-1$ ways to choose a number $x$
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that will replace the number 1.
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There are $n-1$ ways to choose an element $x$
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that replaces the element 1.
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In each such choice, there are two options:
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\textit{Option 1:} We also replace the number $x$
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by the number 1.
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\textit{Option 1:} We also replace the element $x$
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with the element 1.
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After this, the remaining task is to construct
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a derangement for $n-2$ numbers.
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a derangement of $n-2$ elements.
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\textit{Option 2:} We replace the number $x$
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by some other number than 1.
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Now we should construct a derangement
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for $n-1$ numbers, because we can't replace
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the number $x$ with number $1$, and all other
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numbers should be changed.
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\textit{Option 2:} We replace the element $x$
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with some other element than 1.
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Now we have to construct a derangement
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of $n-1$ element, because we cannot replace
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the element $x$ with the element $1$, and all other
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elements should be changed.
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\section{Burnside's lemma}
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\index{Burnside's lemma}
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\key{Burnside's lemma} counts the number of
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combinations so that for each group of
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symmetric combinations, only one representative is counted.
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combinations so that
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only one representative is counted
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for each group of symmetric combinations,
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Burnside's lemma states that the number of
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combinations is
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\[\sum_{k=1}^n \frac{c(k)}{n},\]
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@ -690,7 +690,7 @@ position of a combination,
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and there are $c(k)$ combinations that
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remain unchanged when the $k$th way is applied.
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As an example, let's calculate the number of
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As an example, let us calculate the number of
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necklaces of $n$ pearls,
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where the color of each pearl is
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one of $1,2,\ldots,m$.
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@ -750,12 +750,12 @@ a total of
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necklaces remain the same,
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where $\textrm{gcd}(k,n)$ is the greatest common
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divisor of $k$ and $n$.
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The reason for this is that sequences
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of pearls of size $\textrm{syt}(k,n)$
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The reason for this is that blocks
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of pearls of size $\textrm{gcd}(k,n)$
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will replace each other.
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Thus, according to Burnside's lemma,
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the number of necklaces is
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\[\sum_{i=0}^{n-1} \frac{m^{\textrm{syt}(i,n)}}{n}. \]
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\[\sum_{i=0}^{n-1} \frac{m^{\textrm{gcd}(i,n)}}{n}. \]
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For example, the number of necklaces of length 4
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with 3 colors is
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\[\frac{3^4+3+3^2+3}{4} = 24. \]
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@ -769,7 +769,7 @@ there are $n^{n-2}$ labeled trees
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that contain $n$ nodes.
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The nodes are labeled $1,2,\ldots,n$,
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and two trees are different
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if either their structure or their
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if either their structure or
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labeling is different.
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\begin{samepage}
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@ -829,11 +829,10 @@ be derived using Prüfer codes.
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A \key{Prüfer code} is a sequence of
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$n-2$ numbers that describes a labeled tree.
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The code is calculated by removing $n-2$
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leaves from the tree.
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At each step, we remove the leaf whose number
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is the smallest, and simultaneously
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add the number of its only neighbor to the code.
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The code is constructed by following a process
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that removes $n-2$ leaves from the tree.
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At each step, the leaf with the smallest label is removed,
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and the label of its only neighbor is added to the code.
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For example, the Prüfer code for
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\begin{center}
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@ -856,10 +855,11 @@ For example, the Prüfer code for
|
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is $[4,4,2]$, because we first remove
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node 1, then node 3 and finally node 5.
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We can calculate a Prüfer code for any tree,
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We can construct a Prüfer code for any tree,
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and more importantly,
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the original tree can be constructed
|
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from the Prüfer code.
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Hence, the number of labeled trees equals
|
||||
the number of Prüfer codes that is $n^{n-2}$.
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the original tree can be reconstructed
|
||||
from a Prüfer code.
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Hence, the number of labeled trees
|
||||
of $n$ nodes equals
|
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$n^{n-2}$, the number of Prüfer codes
|
||||
of size $n$.
|
||||
|
|
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Reference in New Issue