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Part 1 Module 6 More counting problems EXERCISE #1 From The FUNDAMENTALIZER, Part 2 Mrs. Plato is going to order supper at a restaurant. She will choose items from each of the following menu categories: Category A: lettuce salad; cole slaw; clam chowder; potato salad; egg salad; vegetable soup; bean soup; macaroni salad. Category B: lamb; grouper fillet; pork chops; prime rib; shrimp; lobster; roast beef; roast pork; fried chicken; baked chicken; broiled salmon. Category C: baked potato; French fries; hash browns; rice; sweet potato; steamed carrots; green beans; pilaf. How many meal combinations are possible, assuming that Mrs. Plato will choose 3 items from Category A, 3 items from Category B, and 2 items from Category C, and no item will be selected more than once? A. 595,056B. 18,627,840C. 258,720D. 43,614,208

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Solution #1 This is a Fundamental Counting Principle problem, but it is more complicated than those from Part 1 Module 6. Here is a summary of Mrs. Platos three decisions. Category A: Choose three distinct items from a list of eight items. Category B: Choose three distinct items from a list of eleven items. Category C: Choose two distinct items from a list of eight items.

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Solution #1, page 2 This is a Fundamental Counting Principle problem, but it is more complicated than those from Unit 3 Module 1. Here are the number of options for each of the three compound decisions. Category A: Choose three distinct items from a list of eight items. C(8,3) = 56 different combinations Category B: Choose three distinct items from a list of eleven items. C(11,3) = 165 different combinations Category C: Choose two distinct items from a list of eight items. C(8,2) = 28 different options According to the Fundamental Counting Principle, the number of outcomes is 56x165x28 = 258,720 Choice C is correct.

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Exercise #2 From The FUNDAMENTALIZER, Part 2 Mr. Moneybags, while out for a stroll, encounters a group of ten children. He has in his pocket four shiny new dimes and three shiny new nickels that he will give to selected children. In how many ways may these coins be distributed among the children, assuming that no child will get more than one coin? A. 25,200B. 3,628,800 C. 4,200D. 604,800

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Solution #2 Mr. Moneybags, while out for a stroll, encounters a group of ten children. He has in his pocket four shiny new dimes and three shiny new nickels that he will give to selected children. In how many ways may these coins be distributed among the children, assuming that no child will get more than one coin? This is a Fundamental Counting Principle problem, involving two compound decisions. Mr. Moneybags must: 1. Choose four children to receive the dimes, and 2. Choose three other children to receive the nickels.

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Solution #2, page 2 Summary: He has ten children to choose from, and Mr. Moneybags must: 1. Choose four children to receive the dimes C(10,4) = 210 different combinations 2. Choose three other children to receive the nickels. These means that he must choose three children from among the six who didnt get dimes. C(6,3) = 20 different combinations According to the Fundamental Counting Principle, the number of outcomes is 210 x 20 = 4,200 The correct choice is C.

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Exercise #3 There are nine waitresses and six busboys employed at the trendy new restaurant The House of Hummus. From among each of these groups The International Brother/Sisterhood of Table Service Workers will select a shop steward and a secretary. How many outcomes are possible? A. 51B. 540 C. 2160D. 102 E. None of these

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Solution #3 This is a Fundamental Counting Principle problem involving two compound decisions. 1. Choose a shop steward and a secretary from among the nine waitresses. 2. Choose a shop steward and a secretary from among the six busboys. Note that each of these two decisions involves permutations, not combinations, because the people being selected are getting different jobs or titles.

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Solution #3, page 2 1. Choose a shop steward and a secretary from among the nine waitresses. P(9,2) = 72 options 2. Choose a shop steward and a secretary from among the six busboys. P(6,2) = 30 options According to the Fundamental Counting Principle, the number of outcomes is 72x30 = 2,160 Choice C is correct.

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Why would we add? In the examples we have looked at so far, we found various numbers, perhaps by using the permutation formula or the combination formula, and then multiplied those numbers. In some situations, however, we may want to add numbers, rather than multiply, at the end of a calculation. The following example suggests why we might add, rather the multiply.

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Why would we add? EXAMPLE Suppose we ask, How many of you are 18 years old? and, by show of hands, we see that there are 25 people who are 18 years of age. Next, suppose we ask, How many of you are 19 years old and, by show of hands, we see that there are 30 people who are 19 years of age. Now we want to use those results to answer the question How many are 18 or 19 years old? It would make no sense to say that the number of people who are 18 or 19 years old is 25x30 = 750.

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Why would we add? If 25 people are 18 years old, and 30 people are 19 years old, then the number of people who are 18 or 19 years old is = 55. This suggests the following fact: To find the number of outcomes in an either…or situation, find the number of of options for each case, and add them. More formally, if A, B are mutually exclusive conditions, then n(A or B) = n(A) + n(B).

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OR means ADD Generally, in counting problems or probability problems, AND means MULTIPLY OR means ADD

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Exercise #3a There are nine waitresses and six busboys employed at the trendy new restaurant The House of Hummus. The International Brother/Sisterhood of Table Service Workers will either select a shop steward and a secretary from among the waitresses, or they will select a shop steward and a secretary from among the busboys. How many outcomes are possible? A. 51B. 540 C. 2160D. 102 E. None of these

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Solution #3a This is an either/or scenario involving two mutually exclusive cases Either Case 1. Choose a shop steward and a secretary from among the nine waitresses. or Case 2. Choose a shop steward and a secretary from among the six busboys. Note that each of these two cases involves permutations, not combinations, because the people being selected are getting different jobs or titles.

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Solution #3a, page 2 Case 1. Choose a shop steward and a secretary from among the nine waitresses. P(9,2) = 72 options Case 2. Choose a shop steward and a secretary from among the six busboys. P(6,2) = 30 options Since we will perform either Case 1, or Case 2, but not both, the number of outcomes is = 102 Choice D is correct.

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EXERCISE #4 A couple is expecting the birth of a baby. If the child is a girl, they will choose her first name and middle name from this list of their favorite girls names: Betty, Beverly, Bernice, Bonita, Barbie. If the child is a boy, they will choose his first name and middle name from this list of their favorite boys names: Biff, Buzz, Barney, Bart, Buddy, Bert. In either case, the child's first name will be different from the middle name. How many two-part names are possible? A. 50B. 600C. 61D. 900

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Solution #4 Girls names: Betty, Beverly, Bernice, Bonita, Barbie. Boys names: Biff, Buzz, Barney, Bart, Buddy, Bert. Any two-part name will be either a girls name or a boys name, so we need to find the number of possible girls names, the number of possible boys names, and then add these two numbers. To form a two-part girls name, there are five options for the first name and then four options for the middle name, so the number of outcomes is 5x4=20 (this is the same as P(5,2)). To form a two-part boys name, there are six options for the first name and then five options for the middle name, so the number of outcomes is 6x5=30 (this is the same as P(6,2)). There are 20 possible two-part girls names, and 30 possible two-part boys names, so the number of two-part names is 20+30=50.

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Exercise #5 The mathematics department is going to hire a new instructor. They want to hire somebody who possesses at least four of the following traits: 1. Honest; 2. Trustworthy; 3. Loyal; 4. Gets along well with others; 5. Good at math; 6. Good handwriting In how many ways is it possible to combine at least four of these traits? A. 360 B. 15 C. 22 D. 48 E. None of these

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Solution #5 This is an either…or situation. There are six traits to choose from. In this context, choosing at least four traits means that we choose either four traits or five traits or six traits. Either: 1. Choose four traits: C(6,4) = 15 or 2. Choose five traits: C(6,5) = 6 or 3. Choose six traits: C(6,6) = = 22 different ways to combine at least four of the six traits.

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Exercise #6 Tonight there are 8 bus boys and 7 dishwashers working at the trendy new restaurant Cap'n Krusto's Crustacean Castle. Because it is a slow night the manager will select either 3 bus boys or 2 dishwashers and send them home early. How many outcomes are possible? A B. 378 C. 77 D E. None of these

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Solution #6 8 bus boys 7 dishwashers Select either 3 bus boys or 2 dishwashers and send them home early. In either…or situations we find the number of options for each case, and then add. The number of ways to choose 3 bus boys is C(8,3) = 56. The number of ways to choose 2 dishwashers is C(7,2) = 21. So, the number of ways to choose 3 bus boys OR 2 dishwashers is = 77. The correct choice is C. Finally, if the problem had said that he was going to choose 3 bus boys and 2 dishwashers, then the answer would be 56x21 = 1176, choice D.

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Exercise #7 The engineers at the Gomermatic Corporation have designed a new model of automatic cat scrubbing machine, and a new model of robotic dog poop scooper. Now it is the Marketing Department's job to come up with catchy, high-tech sounding names for the products. For each product they will randomly generate a three-syllable name, such as Optexa, by choosing one syllable from each of the following categories: First syllable: Apt; Opt; Axt; Emt; Art; Ext. Second syllable: a; y; e. Third syllable: va; xa; ta; ra. How many outcomes are possible, assuming that the two products will not have the same name? A. 5112B. 143 C. 2556D. none of these

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Solution #7 First syllable: Apt; Opt; Axt; Emt; Art; Ext. Second syllable: a; y; e. Third syllable: va; xa; ta; ra. There are two compound decisions. We must 1. Create a three syllable name for the automatic cat scrubbing machine; and 2. Create a three-syllable name for the robotic dog poop scooper. Choosing a three-syllable name for the automatic cat scrubbing machine requires three decisions: Choose the first syllable: 6 options Choose the second syllable: 3 options Choose the third syllable: 4 options. According to the Fundamental Counting Principle, the number of possible names for the automatic cat scrubbing machine is (6)(3)(4) = 72. Having chosen the name for the automatic cat scrubbing machine, there will then be 71 possible names for the robotic dog poop scooper. Thus, the total number of ways to name the two products is (72)(71) = 5112.

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Exercise #8 Among a certain group of 28 mules, 23 are stubborn, 19 are obstinate, and 16 are both stubborn and obstinate. How many are stubborn or obstinate? A. 26 B. 51C. 10D. None of these

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Solution #8 Among a certain group of 28 mules, 23 are stubborn, 19 are obstinate, and 16 are both stubborn and obstinate. How many are stubborn or obstinate? If a counting problem refers to conditions that are not mutually exclusive (typically, a population described in terms of overlapping categories), then we just draw a Venn diagram. The Venn diagram shows that the number who are stubborn or obstinate is = 26 Remember: Or is always inclusive.

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Exercise 9 In the Psychology Department there are 8 faculty members. 2 faculty members will be chosen to design a new course offering and 3 other faculty members will be chosen to hang out in the faculty parking lot to prevent students from parking there. How many different outcomes are possible? A. 560B. 48 C. 240D. None of these.

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SOLUTION #9 This problem involves two dependent, compound decisions, among other things. First, choose two people to design a new course. There are eight people to choose from; since the two people being chosen being assigned to the same task, order doesnt matter. The number of ways to choose two people to design a new course offering is C(8,2) = 28. Next, having chosen two people to design a new course offering there will remain six people to choose from for the job of hanging out in the faculty parking lot to prevent students from parking there, so the number of ways to choose those three people is C(6,3) = 20. Thus, the number of ways to choose two people to design a new course offering AND three other people to hang out in the faculty parking lot to prevent students from parking there is 28x20 = 560. The correct choice is A.

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New problem In the Psychology Department there are 8 faculty members. They will either choose 2 faculty members to design a new course offering or choose 3 faculty members to hang out in the faculty parking lot to prevent students from parking there. How many different outcomes are possible? Solution For this either/or situation, we find the number of outcomes for each of the two cases, and add. If they decide to choose two people to design a new course, the number of outcomes is C(8,2) = 28. On the other hand, if instead they decide to choose three people to hang out in the parking lot, they still have eight people from whom to choose, so the number of outcomes is C(8,3) = 56. Thus, the number of ways to either choose two people to design a new course offering OR choose three people to hang out in the faculty parking lot is = 84.

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