Merge branch 'ex8' into 'master'
Add exercise 8 See merge request oysteikt/v21-ma0301!2
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\documentclass[12pt]{article}
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\usepackage{ntnu}
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\usepackage{ntnu-math}
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\usepackage{ntnu-code}
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\author{Øystein Tveit}
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\title{MA0301 Exercise 8}
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\usepackage{amsthm}
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\usepackage{mathabx}
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\begin{document}
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\ntnuTitle{}
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\break{}
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\begin{excs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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Since there is only one president, the possibilities is the sum of candidates
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\[ 5 + 8 = 13 \]
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\subexc{}
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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
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\[ 5 \cdot 8 = 13 \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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To end up with the amount of possibilities, we have to multiply the amounts of components together
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\[ 4 \cdot 12 \cdot 3 \cdot 2 = 288 \]
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\subexc{}
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This reduces the amount of colors from 4 to 1. Therefore the amount of possibilities is
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\[ 1 \cdot 4 \cdot 3 \cot 2 = 24 \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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Let one bakery item be either pastry or muffins.
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\[ (8 + 6) \cdot (4 + 6 + 1 + 1) = 14 \cdot 12 = 168 \]
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\subexc{}
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\[ 14 \cdot 4 \cdot 6 \cdot 6 = 2016 \]
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\subexc{}
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\[ 8 \cdot 6 \cdot 6 \cdot 1 \cdot 2(14 \cdot 4) = 32256 \]
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\end{subexcs}
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\exc{}
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\[ \nPr{8!}{8!} = \frac{8!}{(8-8)!} = 8! = 40320 \]
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\exc{}
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\begin{subexcs}
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\subexc{}
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\[ \nPr{7}{2} = \frac{7!}{(7-2)!} = \frac{7!}{5!} = 7 \cdot 6 = 42 \]
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\subexc{}
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\[ \nPr{8}{4} = \frac{8!}{(8-4)!} = 8 \cdot 7 \cdot 6 \cdot 5 = 1680 \]
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\subexc{}
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\[ \nPr{10}{7} = \frac{10!}{(10-7)!} = 10 \cdot 9 \cdot \ldots \cdot 4 = 604800 \]
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\subexc{}
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\[ \nPr{12}{3} = \frac{12!}{(12-3)!} = 12 \cdot 11 \cdot 10 = 1320 \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\[ \nCr{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \cdot \cancelto{3}{9} \cdot \cancelto{1}{8} \cdot 7}{1 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4}} = 10 \cdot 3 \cdot 7 = 210 \]
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\subexc{}
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\[ \nCr{12}{7} = \frac{12!}{7!5!} = \frac{\cancelto{1}{12} \cdot 11 \cdot \cancelto{1}{10} \cdot 9 \cdot 8}{1 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4} \cdot \cancel{5}} = 11 \cdot 9 \cdot 8 = 792 \]
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\subexc{}
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\[ \nCr{14}{12} = \frac{14!}{12!2!} = \frac{14 \cdot 13}{2} = 91 \]
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\subexc{}
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\[ \nCr{15}{10} = \frac{15!}{10!5!} = \frac{\cancelto{1}{15} \cdot \cancelto{7}{14} \cdot 13 \cdot \cancelto{3}{12} \cdot 11}{1 \cdot \cancel{2} \cdot \cancel{3} \cdot \cancel{4} \cdot \cancel{5}} = 7 \cdot 13 \cdot 3 \cdot 11 = 3003 \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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If there are no restrictions, the amount of arrangements will be the number of permutations of the books
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\[ \nPr{7}{7} = 7! \]
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In order for the languages to alternate, the 3 books has to be inbetween each of the 4 books
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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.
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Hence the amount of ways to arrange the books will be
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\subexc{}
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\[ \nPr{3,3} \cdot \nPr{4}{4} = 3! \cdot 4! = 144 \]
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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.
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For each of those blocks, there is $\nPr{3}{3}$ arrangements of the C++ books.
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Hence the amount of ways to arrange the books will be
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\subexc{}
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\[ \nPr{3}{3} \cdot \nPr{5}{5} = 3! \cdot 5! = 720 \]
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Here, there are just two blocks. Therefore there is only $\nPr{2}{2}$ permutations of the blocks
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For each of those permutations, there are $\nPr{3}{3}$ ways to arrange the C++ books.
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And for each way to arrange the C++ books, there are $\nPr{4}{4}$ ways to arrange the java books.
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Hence the amount of ways to arrange the books will be
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\subexc{}
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\[ \nPr{2}{2} \cdot \nPr{3}{3} \cdot \nPr{4}{4} = 288 \]
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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Because we don't care if there's a different order that the people were selected, we have to use combinations.
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No restrictions means every combination of 12 in 20 people.
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\[ \nCr{20}{12} = 125970 \]
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\subexc{}
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For every combination of six women, we have every combination of six men.
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\[ \nCr{10}{6} \cdot \nCr{10}{6} = 44100 \]
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\subexc{}
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Here we sum together all the combinations where there is an even number of women for every corresponding combination of men.
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\[ \sum^5_{i=0} \nCr{10}{2i} \cdot \nCr{10}{2(6-i)} = 63090 \]
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\subexc{}
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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.
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Therefore, we sum together all combination products from 7 to 10.
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\[ \sum^{10}_{i=7} \nCr{10}{i} \cdot \nCr{10}{12-i} = 40935 \]
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\subexc{}
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Sum of all combination products from 8 to 10.
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\[ \sum^{10}_{i=8} \nCr{10}{i} \cdot \nCr{10}{12-i} = 10695 \]
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\end{subexcs}
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\exc{}
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In order to solve this task, we will sum together separate cases \\
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Case i) The number only contains one distinct digit from $\{1, 3, 7, 8\}$
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This would be all the permutations of the digits, that is
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\[ \nPr{4}{4} \] \\
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Case ii) The number contains 2 of the digit $3$ and two distinct digits from $\{1, 7, 8\}$
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Here, we start with all the ways we can form a four digit number including two of the digit $3$.
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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)
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Therefore the amount of ways we can write a four digit number including two of the digit $3$ would be
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\[\nCr{4}{2}\]
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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)
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Hence, the amount of ways we can write this number would be
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\[\nCr{4}{2} \cdot \nPr{3}{3}\] \\
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Case iii) The number contains 2 of the digit $7$ and two distinct digits from $\{1, 3, 8\}$
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This is the same as Case ii, just with $7$s instead of $3s$ \\
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Case iv) The number contains 2 of the digit $3$ and 2 of the digit $7$
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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.
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Hence, the amount of ways we can write this number would be
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\[\nCr{4}{2}\]
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In conclusion, the total amount of distinct four digit integers we can make with the digits $1$, $3$, $3$, $7$, $7$, $8$ would be
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\[ \nPr{4}{4} + \nCr{4}{2} \cdot \nPr{3}{3} \cdot 2 + \nCr{4}{2} = 102 \]
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\exc{}
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Here i wrote a program to calculate a modified pascal triangle and print the coefficients of a specific row
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\codeFile{scripts/pascal.py}{python}
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Here are the results:
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\begin{verbatim}
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a.
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1 x^12 y^0
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12 x^11 y^1
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66 x^10 y^2
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220 x^9 y^3
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495 x^8 y^4
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792 x^7 y^5
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924 x^6 y^6
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792 x^5 y^7
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495 x^4 y^8
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220 x^3 y^9
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66 x^2 y^10
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12 x^1 y^11
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1 x^0 y^12
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b.
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1 x^12 y^0
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24 x^11 y^1
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264 x^10 y^2
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1760 x^9 y^3
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7920 x^8 y^4
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25344 x^7 y^5
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59136 x^6 y^6
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101376 x^5 y^7
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126720 x^4 y^8
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112640 x^3 y^9
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67584 x^2 y^10
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24576 x^1 y^11
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4096 x^0 y^12
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c.
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4096 x^12 y^0
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-73728 x^11 y^1
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608256 x^10 y^2
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-3041280 x^9 y^3
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10264320 x^8 y^4
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-24634368 x^7 y^5
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43110144 x^6 y^6
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-55427328 x^5 y^7
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51963120 x^4 y^8
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-34642080 x^3 y^9
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15588936 x^2 y^10
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-4251528 x^1 y^11
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531441 x^0 y^12
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\end{verbatim}
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Which means that the coefficient for $x^9y^3$ for each of the subexercises would be
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\begin{subexcs}
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\subexc{}
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\[ 220 \]
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\subexc{}
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\[ 17601 \]
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\subexc{}
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\[ -3041280 \]
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\end{subexcs}
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\end{excs}
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\end{document}
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@ -0,0 +1,25 @@
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from itertools import permutations, combinations
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def amountOfBooksWhere(fun):
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return len([bookOrder for bookOrder in permutations('jjjjccc') if fun(''.join(bookOrder))])
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# a)
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def no_restriction(books):
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return True
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# b)
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def alternating_books(books):
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return all(books.find(l+l) == -1 for l in set(books))
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# c)
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def all_cs_are_together(books):
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return not books.find('ccc') == -1
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# d)
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def all_cs_js_are_together(books):
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return not (books.find('ccc') == -1 or books.find('jjjj') == -1)
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print(amountOfBooksWhere(no_restriction))
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print(amountOfBooksWhere(alternating_books))
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print(amountOfBooksWhere(all_cs_are_together))
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print(amountOfBooksWhere(all_cs_js_are_together))
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@ -0,0 +1,38 @@
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from math import comb
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def pascal(n, left, right):
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i = 1
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result = [[left, right]]
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while i < n:
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prev_row = result[-1]
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result.append(
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[left ** (i+1)]
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+ [comb(i+1, j+1) * (left ** (len(prev_row)-j-1)) * (right ** (j+1)) for j in range(len(prev_row)-1)]
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+ [right ** (i+1)])
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i += 1
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return result
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def pretty_print_pascal(triangle):
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for i, row in enumerate(triangle):
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print(f'{i+1}.\t{row}')
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def pretty_print_row(row):
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for i, coefficient in enumerate(row):
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print(f'{coefficient}\tx^{len(row)-i-1}\ty^{i}')
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if __name__ == "__main__":
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print('a. ')
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pretty_print_row(pascal(12, 1, 1)[-1])
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print()
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print('b. ')
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pretty_print_row(pascal(12, 1, 2)[-1])
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print()
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print('c. ')
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pretty_print_row(pascal(12, 2, -3)[-1])
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