Add A&W summary. Still WIP for fixing errors and more proofs

semester2/algorithms-and-probability/parts/combinatorics.tex

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+\newsection
+\section{Combinatorics}
+\subsection{Introduction}
+Combinatorics was developed from the willingness of humans to gamble and the fact that everybody wanted to win as much money as possible.
+
+\subsection{Simple counting operations}
+The easiest way to find the best chance of winning is to write down all possible outcomes. This can be very tedious though when the list gets longer.
+
+We can note this all down as a list or as a tree diagram. So-called Venn Diagrams might also help represent the relationship between two sets or events. Essentially a Venn Diagram is a graphical representation of set operations such as $A \cup B$.
+
+
+\subsection{Basic rules of counting}
+\subsubsection{Multiplication rule}
+If one has $n$ possibilities for a first choice and $m$ possibilities for a second choice, then there are a total of $n \cdot m$ possible combinations.
+
+When we think about a task, and we have an \textbf{and} in between e.g. properties, we need to multiply all the options.
+
+\subsubsection{Addition rule}
+If two events are mutually exclusive, the first has $n$ possibilities and the second one has $m$ possibilities, then both events together have $n+m$ possibilities.
+
+When we think about a task, and we have an \textbf{or} in between e.g. properties, then we need to add all the options.
+
+
+\newpage
+\subsection{Factorial}
+\begin{definition}[]{Factorial}
+    The factorial stands for the product of the first $n$ natural numbers where $n \ge 1$. Notation: $!$
+    \[
+        n! = n \cdot (n - 1) \cdot (n - 2) \cdot \ldots \cdot 3 \cdot 2 \cdot 1
+    \]
+    Additionally, $0! = 1$. We read $n!$ as ``\textit{n factorial}''
+\end{definition}
+
+\subsubsection{Operations}
+We can rewrite $n!$ as $n \cdot (n - 1)!$ or $n \cdot (n - 1) \cdot (n - 2)!$ and so on.
+
+It is also possible to write $7 \cdot 6 \cdot 5$ with factorial notation: $\displaystyle \frac{7!}{4!}$, or in other words, for any excerpt of a factorial sequence: \[n \cdot (n - 1) \cdot \ldots \cdot m = \frac{n!}{(m - 1)!}\]
+
+
+\subsection{Permutations}
+\begin{definition}[]{Permutations}
+    A permutation of a group is any possible arrangement of the group's elements in a particular order\\
+
+    \textbf{Permutation rule without repetition:} The number of $n$ \textbf{\textit{distinguishable}} elements is defined as: $n!$
+\end{definition}
+
+
+\subsubsection{Permutation with repetition}
+For $n$ elements $n_1,n_2,\ldots,n_k$ of which some are identical, the number of permutations can be calculated as follows:
+\[
+    p = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}
+\]
+where $n_k$ is the number of times a certain element occurs. 
+As a matter of fact, this rule also applies to permutations without repetition, as each element occurs only once, which means the denominator is $1$, hence $\displaystyle \frac{n!}{(1!)^n} = n!$
+
+\inlineex \smallhspace CANADA has $6$ letters, of which $3$ letters are the same. So the word consists of $3$ A's, which can be arranged in $3!$ different ways, a C, N and D, which can be arranged in $1!$ ways each. Therefore, we have:
+\[
+    \frac{6!}{3!\cdot 1! \cdot 1! \cdot 1!} = \frac{6!}{3!} = 6 \cdot 5 \cdot 4 = 120
+\]
+
+Since $1!$ equals $1$, we can always ignore all elements that occur only once, as they won't influence the final result.
+
+
+\newpage
+\subsection{Variations}
+\begin{definition}[]{Variations}
+    A \textbf{\textit{variation}} is a selection of $k$ elements from a universal set that consists of $n$ \textit{distinguishable} elements.\\
+
+    \textbf{Variation rule without repetition:} The $_n\mbox{P}_k$ function is used to \textit{\textbf{place}} $n$ elements on $k$ places. In a more mathematical definition:
+    The number of different variations consisting of $k$ different elements selected from $n$ distinguishable elements can be calculated as follows:
+
+    \[
+        \frac{n!}{(n - k)!} = _n\mbox{P}_k
+    \]
+\end{definition}
+
+\subsubsection{Variations with repetition}
+If an element can be selected more than once and the order matters, the number of different variations consisting of $k$ elements selected from $n$ distinguishable elements can be calculated using $n^k$
+
+
+
+\subsection{Combinations}
+\begin{definition}[]{Combination}
+    A combination is a selection of $k$ elements from $n$ elements in total without any regard to order or arrangement.
+
+    \textbf{Combination rule without repetition:} \[
+        _n\mbox{C}_k = {n\choose k} = \frac{_n\mbox{P}_k}{k!} = \frac{n!}{(n - k)! \cdot k!}
+    \]
+\end{definition}
+
+\subsubsection{Combination with repetition}
+In general the question to ask for combinations is, in how many ways can I distribute $k$ objects among $n$ elements?
+\[
+    _{n + k - 1}\mbox{C}_k = {n + k - 1\choose k} = \frac{(n + k - 1)!}{k!(n - 1)!}
+\]
+
+
+
+\subsection{Binomial Expansion}
+\label{sec:binomial-expansion}
+Binomial expansion is usually quite hard, but it can be much easier than it first seems. The first term of the expression of $(a + b)^n$ is always $1 a^n b^0$. Using the formula for combination without repetition, we can find the coefficients of each element:
+
+\begin{center}
+    \includegraphics[width=0.6\linewidth]{./assets/binomialExpansion.png}
+\end{center}
+
+This theory is based on the Pascal's Triangle and the numbers of row $n$ correspond to the coefficients of each element of the expanded term.
+
+We can calculate the coefficient of each part of the expanded term $k$ with combinatorics as follows: $\displaystyle {n\choose k}$
+
+\begin{formula}[]{Binomial Expansion}
+    \textbf{\textit{\underbar{In general:}}}
+    \[
+        (a + b)^n = 1a^nb^0 + {n\choose 1} a^{n-1}b^{1} + {n\choose 2} a^{n-2}b^{2} + \ldots + {n\choose n - 1} a^{1}b^{n - 1} + {n\choose n} a^{0}b^{n}
+    \]
+\end{formula}
+
+
+\subsection{Overview}
+\includegraphics[width=1\linewidth]{./assets/overview.png}