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luku22.tex
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luku22.tex
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@ -29,7 +29,7 @@ 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 problem, we can define a function $f(n)$
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that counts the number of representations for $n$.
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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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@ -42,7 +42,7 @@ can be recursively calculated as follows:
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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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select the last number in the sum.
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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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@ -63,7 +63,7 @@ It turns out that the function also has a closed-form formula
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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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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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@ -369,8 +369,8 @@ 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 expression \texttt{()} is valid.
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\item If a expression $A$ is valid,
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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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@ -397,7 +397,7 @@ 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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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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@ -559,12 +559,12 @@ that corresponds to the area of the region
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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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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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of sets is larger.
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When there are three sets, the inclusio-exclusion formula is
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When there are three sets, the inclusion-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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@ -594,7 +594,7 @@ If the intersection contains an odd number of sets,
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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 can also be used
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Note that there are similar formulas
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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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@ -678,10 +678,10 @@ elements should be changed.
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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
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\key{Burnside's lemma} can be used to count
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the number of 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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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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