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LSU MATH 2020 - Lecture Notes

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Math2020 Solutions to question 7 of final guide(1) (a) How many ways are there to select three different pieces of fruit froma bowl of 5 different fruits?SOLUTION: Ways to choose 3 from 5 is53=5 × 4 × 31 × 2 × 3= 10(b) You go to the supermarket to buy three peices of fruit, which may ormay not all be the same. The supermarket sells 5 different kinds offruit. How many different possibilities are there for what you comehome with?SOLUTION: Ways to choose 3 from 5 types of things, order doesn’tmatter, repeats are allowed:3 + 5 − 15 − 1=73=7 × 6 × 51 × 2 × 3= 35(c) You are painting the windows frames on the front of a house. Thereare three windows and 5 choices of colors. How many different wayscould you paint them, if there is no requirement that they have to bedifferent colors from each other?SOLUTION: 3 choices from 5 possible types. Order does matter. Re-peats are allowed:5 × 5 × 5 = 125(d) You are painting the windows frames on the front of a house. Thereare three windows and 5 choices of colors. How many different wayscould you paint them, if there is a requirement that they all have tobe different colors from each other?SOLUTION: 3 choices from 5 possible types. Order does matter. Re-peats are not allowed:P (5, 3) = 5 × 4 × 3 = 60(e) You have 5 pieces of fruit which you are lining up on your kitchencounter. Your fruits are: 2 (identical) apples, 2 (identical) bananas,and 1 orange. How many different ways can you line them up?SOLUTION: Ways of ordering 5 things of three types. 2 of type 1, 2of type 2, 1 of type 3. Order matters.5!2!2!1!=5 × 4 × 3 × 2 × 12 × 2 × 1= 30(f) You have 5 pieces of fruit which you are lining up on your kitchencounter. Your fruits are: 2 (identical) apples and 3 (identical) bananas.How many different ways can you line them up?SOLUTION: Ways of ordering 5 things of two types. 2 of type 1, 3 oftype 2. Order matters.5!2!3!=5 × 4 × 3 × 2 × 12 × 6= 1012(g) You have 5 pieces of fruit of which you are lining up three on yourkitchen counter (you will eat the remaining 2). Your fruits are: 2(identical) apples, 2 (identical) bananas, and 1 orange. How manydifferent ways can you line three of them them up?SOLUTION:Choose 3 from 5, order does matter; some objects are identical. 2 oftype 1, 2 of type 2, 1 of type 3.Suppose we lined up all five. Then there are5!2!2!1!= 30ways, as in question (e). We can eat the last two things in the line.But when we do this, we count some arrangements twice. E.g., writingA for apple, B for banana, O for orange, we haveAABOBAABBOboth giveAABwhen the last two fruits are removed. Any arrangement with the lasttwo fruits different will lead to an arrangement of three fruits beingcounted twice.How many ways are there to arrange the fruits with the last two dif-ferent?If X is the number with the last two different, and Y the number withthe last two the same, then X + Y is the total number, so X +Y = 30,and X = 30 − Y .If the last two are the same, they are either two apples, or two bananas.If we have two apples at the end, there are two bananas and one orangeleft to arrange, so there are3!2!1!= 3ways to arrange the remaining fruit. If we have two bananas at theend, there are similarly3!2!1!= 3ways to arrange the remaining fruit.So there are 3+3 = 6 ways to arrange fruit with the last two the same.So there are 30 − 6 = 24 ways to arrange with the last two different.Each of these 24 ways gives a double count of the first three piecesof fruit. So there are 12 ways to arrange the fruits with two differentpieces of fruit left over. There are 6 ways with the same fruit left.So there are 6 + 12 = 18 ways to arrange three of the 5 fruits.ANSWER: 18.(h) You go to the cinema with three friends. If you all sit in four adjacentseats, how many different ways can you sit together?SOLUTION: Ways to order 4 different things:P (4, 4) = 4! = 243(i) You go to the cinema with three friends. Two of them must sit inadjacent seats to each other, but noone else cares about seating. Howmany ways can you sit together in four adjacent seats?SOLUTION: Ways to order 4 things, but two must be adjacent. Thereare 24 total ways without this requirement, as in previous question.If two people are always together, we can think of them as one unit,and just choose an ordering of three things P (3, 3) = 3! = 6. But foreach way of putting those two people together, there are two ways,since they can swap seats with each other and still be together. Sothere are2P (3, 3) = 2 × 6 = 12ways(j) You go to the cinema with three friends. Two of them will not sit nextto each other. How many ways are there for you all to sit together infour adjacent seats?SOLUTION: There are 24 total ways, as in question (h). There are 12ways where these two people are together, as in question (i). So thisleaves 24 − 12 ways where they are not together.ANSWER:


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