\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}