\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}
\usepackage{ntnu-code}

\author{Øystein Tveit}
\title{MA0301 Exercise 8}

\usepackage{amsthm}
\usepackage{mathabx}

\begin{document}
  \ntnuTitle{}
  \break{}
  
  \begin{excs}

  \exc{}
  \begin{subexcs}
    \subexc{}
    Since there is only one president, the possibilities is the sum of candidates
    \[ 5 + 8 = 13 \]

    \subexc{}
    For every candidate from one party there is all the candidates of the other party to be compared to. Therefore the amount of possibilities is
    \[ 5 \cdot 8 = 13 \]

  \end{subexcs}

  \exc{}
  \begin{subexcs}
    \subexc{}
    To end up with the amount of possibilities, we have to multiply the amounts of components together
    \[ 4 \cdot 12 \cdot 3 \cdot 2 = 288 \]

    \subexc{}
    This reduces the amount of colors from 4 to 1. Therefore the amount of possibilities is
    \[ 1 \cdot 4 \cdot 3 \cot 2 = 24 \]

  \end{subexcs}

  \exc{}
  \begin{subexcs}
    \subexc{}
    Let one bakery item be either pastry or muffins.
    \[ (8 + 6) \cdot (4 + 6 + 1 + 1) = 14 \cdot 12 = 168 \]

    \subexc{}
    \[ 14 \cdot 4 \cdot 6 \cdot 6 = 2016 \]

    \subexc{}
    \[ 8 \cdot 6 \cdot 6 \cdot 1 \cdot 2(14 \cdot 4) = 32256 \]

  \end{subexcs}

  \exc{}

  \[ \nPr{8!}{8!} = \frac{8!}{(8-8)!} = 8! = 40320 \]

  \exc{}
  \begin{subexcs}
    \subexc{}
    \[ \nPr{7}{2} = \frac{7!}{(7-2)!} = \frac{7!}{5!} = 7 \cdot 6 = 42 \]

    \subexc{}
    \[ \nPr{8}{4} = \frac{8!}{(8-4)!} = 8 \cdot 7 \cdot 6 \cdot 5 = 1680 \]

    \subexc{}
    \[ \nPr{10}{7} = \frac{10!}{(10-7)!} = 10 \cdot 9 \cdot \ldots \cdot 4 = 604800 \]

    \subexc{}
    \[ \nPr{12}{3} = \frac{12!}{(12-3)!} = 12 \cdot 11 \cdot 10 = 1320 \]

  \end{subexcs}

  \exc{}
  \begin{subexcs}
    \subexc{}
    \[ \nCr{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \cdot \cancelto{3}{9} \cdot \cancelto{1}{8} \cdot 7}{1 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4}} = 10 \cdot 3 \cdot 7 = 210 \]

    \subexc{}
    \[ \nCr{12}{7} = \frac{12!}{7!5!} = \frac{\cancelto{1}{12} \cdot 11 \cdot \cancelto{1}{10} \cdot 9 \cdot 8}{1 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4} \cdot \cancel{5}} = 11 \cdot 9 \cdot 8 = 792 \]

    \subexc{}
    \[ \nCr{14}{12} = \frac{14!}{12!2!} = \frac{14 \cdot 13}{2} = 91 \]

    \subexc{}
    \[ \nCr{15}{10} = \frac{15!}{10!5!} = \frac{\cancelto{1}{15} \cdot \cancelto{7}{14} \cdot 13 \cdot \cancelto{3}{12} \cdot 11}{1 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4} \cdot \cancel{5}} = 7 \cdot 13 \cdot 3 \cdot 11 = 3003 \]

  \end{subexcs}

  \exc{}
  \begin{subexcs}
    \subexc{}
    
    If there are no restrictions, the amount of arrangements will be the number of permutations of the books

    \[ \nPr{7}{7} = 7! \]

    In order for the languages to alternate, the 3 books has to be inbetween each of the 4 books

    There are $\nPr{3}{3}$ permutations of the C++ books. For each of those permutations, there are $\nPr{4}{4}$ permutations of the Java books.

    Hence the amount of ways to arrange the books will be

    \subexc{}
    \[ \nPr{3,3} \cdot \nPr{4}{4} = 3! \cdot 4! = 144 \]

    Because all the C++ books has to be together, we can think of them like one "block" in the permutations of 5 blocks where four of the blocks are Java books.

    For each of those blocks, there is $\nPr{3}{3}$ arrangements of the C++ books.

    Hence the amount of ways to arrange the books will be

    \subexc{}
    \[ \nPr{3}{3} \cdot \nPr{5}{5} = 3! \cdot 5! = 720 \]

    Here, there are just two blocks. Therefore there is only $\nPr{2}{2}$ permutations of the blocks

    For each of those permutations, there are $\nPr{3}{3}$ ways to arrange the C++ books.

    And for each way to arrange the C++ books, there are $\nPr{4}{4}$ ways to arrange the java books.

    Hence the amount of ways to arrange the books will be

    \subexc{}
    \[ \nPr{2}{2} \cdot \nPr{3}{3} \cdot \nPr{4}{4} = 288 \]

  \end{subexcs}

  \exc{}
  \begin{subexcs}
    \subexc{}
    Because we don't care if there's a different order that the people were selected, we have to use combinations.

    No restrictions means every combination of 12 in 20 people.

    \[ \nCr{20}{12} = 125970 \]

    \subexc{}
    For every combination of six women, we have every combination of six men.
    \[ \nCr{10}{6} \cdot \nCr{10}{6} = 44100 \]

    \subexc{}
    Here we sum together all the combinations where there is an even number of women for every corresponding combination of men.
    \[ \sum^5_{i=0} \nCr{10}{2i} \cdot \nCr{10}{2(6-i)} = 63090 \]

    \subexc{}
    In order for the selection to contain more women than men, the amount of women has to be 7 so that the amount of men is 5.

    Therefore, we sum together all combination products from 7 to 10.
    \[ \sum^{10}_{i=7} \nCr{10}{i} \cdot \nCr{10}{12-i} = 40935 \]

    \subexc{}
    Sum of all combination products from 8 to 10.
    \[ \sum^{10}_{i=8} \nCr{10}{i} \cdot \nCr{10}{12-i} = 10695 \]

  \end{subexcs}

  \exc{}

  In order to solve this task, we will sum together separate cases \\

  Case i) The number only contains one distinct digit from $\{1, 3, 7, 8\}$

  This would be all the permutations of the digits, that is
  
  \[ \nPr{4}{4} \] \\

  Case ii) The number contains 2 of the digit $3$ and two distinct digits from $\{1, 7, 8\}$

  Here, we start with all the ways we can form a four digit number including two of the digit $3$.

  Since we don't care what order the $3$s are in, we want the combinations (for example, $x{\color{red}3}{\color{ForestGreen}3}x$ and $x{\color{ForestGreen}3}{\color{red}3}x$ are the same)

  Therefore the amount of ways we can write a four digit number including two of the digit $3$ would be
  
  \[\nCr{4}{2}\]

  For each of those ways to write the number, there are all the permutations of the remaining digits ways to construct a number (here order does matter since the digits are distinct)

  Hence, the amount of ways we can write this number would be

  \[\nCr{4}{2} \cdot \nPr{3}{3}\] \\

  Case iii) The number contains 2 of the digit $7$ and two distinct digits from $\{1, 3, 8\}$

  This is the same as Case ii, just with $7$s instead of $3s$ \\

  Case iv) The number contains 2 of the digit $3$ and 2 of the digit $7$

  In this case, the amount of combinations would be the same as the amount of the ways we can write a four digit number with two fixed numbers.

  Hence, the amount of ways we can write this number would be

  \[\nCr{4}{2}\]

  In conclusion, the total amount of distinct four digit integers we can make with the digits $1$, $3$, $3$, $7$, $7$, $8$ would be

  \[ \nPr{4}{4} + \nCr{4}{2} \cdot \nPr{3}{3} \cdot 2 + \nCr{4}{2} = 102 \]

  \exc{}

  Here i wrote a program to calculate a modified pascal triangle and print the coefficients of a specific row

  \codeFile{scripts/pascal.py}{python}

  Here are the results:

  \begin{verbatim}
    a. 
    1       x^12    y^0
    12      x^11    y^1
    66      x^10    y^2
    220     x^9     y^3
    495     x^8     y^4
    792     x^7     y^5
    924     x^6     y^6
    792     x^5     y^7
    495     x^4     y^8
    220     x^3     y^9
    66      x^2     y^10
    12      x^1     y^11
    1       x^0     y^12
    
    b. 
    1       x^12    y^0
    24      x^11    y^1
    264     x^10    y^2
    1760    x^9     y^3
    7920    x^8     y^4
    25344   x^7     y^5
    59136   x^6     y^6
    101376  x^5     y^7
    126720  x^4     y^8
    112640  x^3     y^9
    67584   x^2     y^10
    24576   x^1     y^11
    4096    x^0     y^12
    
    c. 
    4096    x^12    y^0
    -73728  x^11    y^1
    608256  x^10    y^2
    -3041280        x^9     y^3
    10264320        x^8     y^4
    -24634368       x^7     y^5
    43110144        x^6     y^6
    -55427328       x^5     y^7
    51963120        x^4     y^8
    -34642080       x^3     y^9
    15588936        x^2     y^10
    -4251528        x^1     y^11
    531441  x^0     y^12
  \end{verbatim}

  Which means that the coefficient for $x^9y^3$ for each of the subexercises would be

  \begin{subexcs}
    \subexc{}
    \[ 220 \]

    \subexc{}
    \[ 17601 \]

    \subexc{}
    \[ -3041280 \]
    
  \end{subexcs}

  \end{excs}
\end{document}