Name (printed):Student ID #:Section # (or TA’s:name and time)CMSC 250 Quiz #12 ANSWERS Wed., Apr. 21, 2004Write all answers legibly in the space provided. The number of points p os sible for each question is indicatedin square brackets – the total number of points on the quiz is 30, and you will have exactly 15 minutes tocomplete this quiz. You may not use calculators, textbooks or any other aids during this quiz.Note: you only need to take the answer as far as a mathematical expression including addition, sub-traction, multiplication, division, exponents and factorials.1. [15 pnts.] Assume you have fruit to sell in your fruitstand. You have 7 apples, 4 oranges and 6bananas. Fruit of the same type are indistinguishable from each other.You make a fruit basket to deliver to a friend at the hospital. It has 5 pieces of fruit in it answerthe following questions about that fruit basket assuming you filled the fruit basket at random (notthinking about what the friend likes or what would look good in the basket). Assume the order youput the fruit into the basket does not make the basket different.a. How many different baskets can be created with the criteria above?(5+(3−1))!5!(3−1!− 1 =7!5!2!− 1EXPLANATION: We need to distribute the 5 indistinguishable x’s into the three categoriesknown as “apples”, “oranges” and “bananas”. The x’s are indistinguishable from each otherand the bars separating the categories are indistinguishable from each other. Once we have thenumber of ways to distribute the five x’s among the three categories, we must subtract the onethat is not possible bec ause we can’t have a basket that is just oranges since the basket has 5fruits and there are only 4 oranges.b. What is the probability that all of the fruit in the basket are apples?(75)(175)=7!5!2!17!5!12!=7!12!2!17!EXPLANATION: We can not use the number from part “a” since all of those baskets do nothave equal probability. So we need to calculate as our e vent space “how many ways there areto select the 5 apples from the 7 available” and for our sample space “how many ways to select5 fruits from the 17 available”. The probability is then the size of the event space over the sizeof the sample space.c. Assuming you must have at least one of each type of fruit in the basket, how many differentbaskets could be formed?(2+(3−1))!2!(3−1)!=4!2!2!= 6EXPLANATION: We put the 3 pieces of fruit (one of each type into the basket first - then theyare not considered when calculating the number of baskets. Now we only have the other twopieces of fruit to decide upon. So we are distributing 2 X’s among the three categories and usethe formula shown ab ove. Or you can just partition these possibilities - partition into thosewhere the two fruits are of the same type and those where the two fruits are of different types.To calculate where the two fruits are the same type: there are31= 3 ways to select that onefruit. To calculate where the two fruits are of different types: there are32= 3 ways to selectthe two fruits. Since this was a partition you then add the two answers and get 3 + 3 = 6.d. Assume your friend received 2 baskets where the first basket has has 3 apples and 2 orangesand the second basket has 2 apples and 3 bananas. Your friend lines up the fruit in a straitline to decide in what order to eat them. How many different linear arrangements can he makeassuming the fruit of the same type is indistinguishable?1055322=10!5!3!2!= 2520EXPLANATION: There are 10 things to put in a line, but there are subgroups of indistinguish-able things where the subgroups are of size 5, 3 and 2. Or, to look at it a different way: Weneed to select the 5 places where the apples will be placed in the line that is 10 long, then weneed to select the 3 places where the bananas will be placed from the 5 remianing slots, thenwe need to select the 2 places from the 2 remaining s lots for the oranges.2. [15 pnts.] Assume you have 27 indistinguishable chocolate candy bars and 5 distinguishable friends.Your friends want you to share the candy bars with them. Each of the questions is separate from theothers - any conditions mentioned in one question do not carry to any of the others.Note: you only need to take the answer as far as a mathematical expression including addition,subtraction, multiplication, division, exponents and factorials.a. How many ways can you distribute the candy bars among yourself and your 5 frineds?(27+(6−1))!27!(6−1)!=3227!5!=3227EXPLANATION: we need to position the 27 indistinguishable candy bars into the 6 categories(the people receiving the candy bars).b. You don’t want any of your friends (or yourself) to be left out of the candy bar eating so yourecalculate - how many ways are there to distribute the candy bars among yourself and yourfive friends assuming you want to be sure that each person gets at least two candy bars?(15+(6−1))!15!(6−1)!=20!15!5!EXPLANATION: Since 27 − 12 = 15 and there are 12 of the candy bars already distributed,we need to position the 15 remaining candy bars in the 6 categories.c. You decide these candy bars are small so instead of each person receiving at least two as men-tioned in the previous question, you want to be sure you and each of your friends each receiveat least 5 candy bars. Now , how many ways to distribute the candy bars?0EXPLANATION: Since there are only 27 candy bars, you can not give 5 to each of the 6 peoplebecause that would be 30 candy bars (more than you actually have.d. Bill one of the friends is allergic to chocolate, so you need to make sure he does not receiveany of the candy bars. Now how many ways are there to distribute them? (remember ignorerestrictions in previous questions)(27+(5−1))!27!(5−1)!=31!27!4!EXPLANATION: Since B ill is no longer one of the categories that can rece ive candy bars, youonly have 5 categories rather than 6.e. Since Bill is allergic to chocolate and can’t eat the candy bars, you will add 12 peppermintcandycanes to the candy you are distributing. How many ways can you distribute the candyassuming Bill can not eat the chocolate candy bars, but he can eat the peppermint candycanes?(27+(5−1))!27!(5−1)!∗(12+(6−1))!12!(6−1)!=31!27!4!∗17!12!5!EXPLANATION: You must distribute the candy bars to the 5 people who can receive themand then distribute the candycanes to the 6 people who can receive them. Since this is a twostep process where both steps
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