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\title{Calculus en Kansrekenen - assignment 6}
\author{Camil Staps\\\small{s4498062, group Bram}}

\begin{document}

\maketitle
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\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}

