Assignment 6

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+\documentclass[10pt,a4paper]{article}
+
+\usepackage[utf8]{inputenc}
+\usepackage[margin=2cm]{geometry}
+
+\usepackage{enumitem}
+\setenumerate[1]{label=\arabic*.}
+\setenumerate[2]{label=(\alph*)}
+
+% textcomp package is not available everywhere, and we only need the Copyright symbol
+% taken from http://tex.stackexchange.com/a/1677/23992
+\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}}
+
+\usepackage{fancyhdr}
+\renewcommand{\headrulewidth}{0pt}
+\renewcommand{\footrulewidth}{0pt}
+\fancyhead{}
+\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps}
+\pagestyle{fancy}
+
+\usepackage{amsmath}
+\usepackage{amsthm}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\newcommand*\diff{\mathop{}\!\mathrm{d}}
+\DeclareMathAlphabet{\pazocal}{OMS}{zplm}{m}{n}
+
+\parindent0pt
+
+\title{Calculus en Kansrekenen - assignment 6}
+\author{Camil Staps\\\small{s4498062, group Bram}}
+
+\begin{document}
+
+\maketitle
+\thispagestyle{fancy}
+
+\begin{enumerate}
+    \item \begin{enumerate}
+            \item She first chooses one of five flowers and then for every flower she can choose one of four pots. That gives $5\cdot4=20$ possibilities.
+            \item She can choose one combination out of twenty, and then another but not the same out of nineteen. However, we count every combination twice (``$a-b$'' and ``$b-a$''), so we have to take $20\cdot19/2=190$.
+        \end{enumerate}
+
+    \item We first pick the five persons, with ${8\choose5}$ possibilities. Then for every combination we have $5!$ possible permutation. This gives ${8\choose5}\cdot5!=806400$. %todo
+
+    \item \begin{enumerate}
+            \item For the diamonds we have $13$ possibilities out of $52$, for the clubs $13$ out of $51$ and for the hearts $13$ out of $50$. That gives $\frac{13}{52}\cdot\frac{13}{51}\cdot\frac{13}{50}=\frac{13^2}{4\cdot50\cdot51} = \frac{169}{10200}$.
+            \item For the hearts we have $13$ out of $52$. Then for the second card we have $26$ out of $51$. The third card must match the suit of the second, that gives $12$ out of $50$. In total that gives $\frac{13}{52}\cdot\frac{26}{51}\cdot\frac{12}{50} = \frac{13}{425}$.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item We choose the first choose the last digit, out of four ($2,4,6,8$) possibilities. Then, the first four digits can be choosen from $8,7,6$ and $5$ remaining digits. That gives $\frac{8!}{4!} = 1680$ numbers.
+            \item We choose two odd digits with $\frac{5\cdot4}{2}=10$ possibilities. Then we choose the even digits with $\frac{4\cdot3\cdot2}{3!} = 4$ possibilities. Then we can permute these digits in $5!$ possible ways. This gives $10\cdot4\cdot5! = 4800$ possibilities.
+            \item \begin{enumerate}[label=(\alph*)]
+                \item The first four digits aren't restricted. The last digit has four possibilities ($2,4,6,8$). That gives $9^4\cdot4=26244$ numbers.
+                \item Let's first choose the two odd digits, with $\frac{5^2}{2!}$ possibilities, and then the even digits, with $\frac{4^3}{3!}$ possibilities. Then we can permute these digits in $5!$ ways. That gives $\frac{5^2\cdot4^3\cdot5!}{2!\cdot3!} = 16000$ numbers.
+                \end{enumerate}
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item $26^4\cdot10^2 = 45697600$ possible license plates.
+            \item In the range DAA-00-A till FZZ-99-Z would be $3\cdot26^2\cdot10^2\cdot26 = 5272800$ license plates. In the range DAA-00-A till DAZ-99-Z would be $26\cdot10^2\cdot26 = 67600$ license plates. In the range DBA-00-A till DBA-99-Z would be $10^2\cdot26=2600$ license plates. In the range DBB-00-A till DBB-00-Z would be $26$ license plates and in the range DBB-01-A till DBB-01-B are $2$. In total that gives us $5272800 - 67600 - 2600 - 26 - 2 = 5202572$ license plates.
+            \item DFA-78-V is within the range, so it belongs to a moped. DAF-78-A is not in the range, so it belongs to a car. DCF-78-V is in the range, so it belongs to a moped.
+            \item That would be $45797600 - 5202572 = 40595028$ cars.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item $(x-7)^3 = (x-7)(x^2-14x+49) = x^3 -21x^2 + 147x - 343$.
+            \item $(x+2y)^4 = (x^2+4xy+4y^2)^2 = x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4$.
+            \item $(x^3-3)^4 = (x^6 - 6x^3 + 9)^2 = x^{12} -12x^{9} + 54x^6 - 108x^3 + 81$.
+        \end{enumerate}
+
+    \item The first player has to shake $10$ hands, the second $9$, and so on. This gives $\sum_{i=0}^{10} i = 10\cdot\frac{10+1}{2} = 55$.
+
+    \item We first pick three consonants, this gives $4\cdot3\cdot2/3! = 4$ possibilities. Then we pick three vowels with $5\cdot4\cdot3 / 3! = 10$ possibilities. Those $6$ letters have $6!$ permutations, so that gives $4\cdot10\cdot6! = 28800$ different words.
+
+    \item There are ${4+3+3\choose5} = 252$ possibilities to form a five teacher committee. However, ${4+3\choose5} = 21$ of these are formed with only math and English teachers. Therefore, we can form $252-21=231$ committees with at least one IT teacher.
+
+    \item We consider $r$ people. The first person we consider may be born anywhere. The second and further people must be born in the same province, which has a probability $\frac1{12}$ for every not-first person. This gives a probability of $\frac1{12^(r-1)}$, i.e. $1, \frac1{12}, \frac1{144}, \frac1{1728}, \frac1{20736}, \frac1{248832}$ for the values of $r$ listed.
+
+    \item \begin{enumerate}
+            \item $S=\{TT,TH,HT,HH\}$, where the first letter refers to the first throw and the second letter to the second throw, and $T$ stands for tails and $H$ for heads.
+            \item \begin{enumerate}
+                    \item $\{TH,HT\}$.
+                    \item $\{TH,HT,HH\}$.
+                    \item $\{HT,HH\}$.
+                \end{enumerate}
+            \item $P_1 : \pazocal P(S) \to [0,1]$ with $P_1(A) = |A|\cdot\frac14$.
+        \end{enumerate}
+
+    \item \begin{proof}
+            We know $P(S) = 1$. Furthermore, $S$ and $\emptyset$ are mutually exclusive events since $S\cap\emptyset=\emptyset$. Therefore, $P(S\cup\emptyset) = P(S) + P(\emptyset)$. But we know $P(S\cup\emptyset)=P(S)=1$. This then reduces to $1 = 1 + P(\emptyset)$, which means $P(\emptyset)=0$.
+        \end{proof}
+
+    \setcounter{enumi}{13}
+
+    \item \begin{enumerate}
+            \item The probability that we have an Ace on first draw but not on second draw.
+            \item The probability that we have an Ace on first draw or on second draw (or both).
+            \item The porbability that we don't draw an Ace.
+            \item The probability that we draw an Ace on first or third draw (or both), but not on second draw.
+        \end{enumerate}
+
+\end{enumerate}
+
+\end{document}
+