Add exercise 8

exercise8/main.tex

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+\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}

exercise8/scripts/combination-bruteforce.py

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+from itertools import permutations, combinations
+
+def amountOfBooksWhere(fun):
+  return len([bookOrder for bookOrder in permutations('jjjjccc') if fun(''.join(bookOrder))])
+
+# a)
+def no_restriction(books):
+  return True
+
+# b)
+def alternating_books(books):
+  return all(books.find(l+l) == -1 for l in set(books))
+
+# c)
+def all_cs_are_together(books):
+  return not books.find('ccc') == -1
+
+# d)
+def all_cs_js_are_together(books):
+  return not (books.find('ccc') == -1 or books.find('jjjj') == -1)
+
+print(amountOfBooksWhere(no_restriction))
+print(amountOfBooksWhere(alternating_books))
+print(amountOfBooksWhere(all_cs_are_together))
+print(amountOfBooksWhere(all_cs_js_are_together))

exercise8/scripts/pascal.py

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+from math import comb
+
+def pascal(n, left, right):
+  i = 1
+  result = [[left, right]]
+
+  while i < n:
+    prev_row = result[-1]
+    result.append(
+      [left ** (i+1)]
+    + [comb(i+1, j+1) * (left ** (len(prev_row)-j-1)) * (right ** (j+1)) for j in range(len(prev_row)-1)]
+    + [right ** (i+1)])
+
+    i += 1
+
+  return result
+
+def pretty_print_pascal(triangle):
+  for i, row in enumerate(triangle):
+    print(f'{i+1}.\t{row}')
+
+def pretty_print_row(row):
+  for i, coefficient in enumerate(row):
+    print(f'{coefficient}\tx^{len(row)-i-1}\ty^{i}')
+
+
+if __name__ == "__main__":
+
+  print('a. ')
+  pretty_print_row(pascal(12, 1, 1)[-1])
+  print()
+
+  print('b. ')
+  pretty_print_row(pascal(12, 1, 2)[-1])
+  print()
+
+  print('c. ')
+  pretty_print_row(pascal(12, 2, -3)[-1])