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Part 1 Module 6 More counting problems

In Part 1 Module 6, we see counting problems is which two or more simpler problems, like those from Part 1 Modules 4 and 5, are combined into a single, more complicated problem. The following guidelines are usually important:

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Part 1 Module 6 More counting problems

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  1. In Part 1 Module 6, we see counting problems is which two or more simpler problems, like those from Part 1 Modules 4 and 5, are combined into a single, more complicated problem. The following guidelines are usually important: Suppose A and B are tasks, where n(A) is the number of ways to perform Task A, and n(B) is the number of ways to perform Task B. Then: 1. The number of ways to perform both A and B = n(A) × n(B); 2. The number of ways to perform either A or B = n(A) + n(B). Part 1 Module 6 More counting problems

  2. In general, in counting problems and in probability, “And” means “multiply” “Or” means “add” n(A and B) = n(A) × n(B); N(A or B) = n(A) + n(B) when A, B are mutually exclusive. And, Or

  3. EXERCISE #1 Tonight there are 6 dishwashers and 5 servers working at the trendy new restaurant I Definitely Believe It's Tofu. Because it is a slow night the manager will select 2 dishwashers and 4 servers and send them home early.How many outcomes are possible?A. 75 B. 20 C. 3600 D. None of these. EXERCISE #2 Tonight there are 6 dishwashers and 5 servers working at the trendy new restaurant I Definitely Believe It's Tofu. Because it is a slow night the manager will select either 2 dishwashers or 4 servers and send them home early.How many outcomes are possible?A. 75 B. 20 C. 3600 D. None of these. Part 1 Module 6 More counting problems

  4. EXERCISE #1 Tonight there are 6 dishwashers and 5 servers working at the trendy new restaurant I Definitely Believe It's Tofu. Because it is a slow night the manager will select 2 dishwashers and 4 servers and send them home early.How many outcomes are possible?A. 75 B. 20 C. 3600 D. None of these. The number of ways to choose 2 dishwashers is C(6, 2) = 15. The number of ways to choose 4 servers is C(5, 4) = 5. So, the number of ways to choose 2 dishwashers and 4 servers is 15×5 = 75 Solutions

  5. EXERCISE #2 Tonight there are 6 dishwashers and 5 servers working at the trendy new restaurant I Definitely Believe It's Tofu. Because it is a slow night the manager will select either 2 dishwashers or 4 servers and send them home early.How many outcomes are possible?A. 75 B. 20 C. 3600 D. None of these. The number of ways to choose 2 dishwashers is C(6, 2) = 15. The number of ways to choose 4 servers is C(5, 4) = 5. So, the number of ways to choose either 2 dishwashers or 4 servers is 15+5 = 20 Solutions

  6. From The FUNDAMENTALIZER, Part 2 www.math.fsu.edu/~wooland/count/count17.html 1. 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. 25,200 B. 3,628,800 C. 4,200 D. 604,800 2. Suddenly, he feels too cheap to give away all his coins. Instead, he will choose either four kids to receive the dimes or three kids to receive the nickels. In how many ways may the coins be distributed among the children, assuming that no child will get more than one coin? Exercises

  7. From The FUNDAMENTALIZER, Part 2 www.math.fsu.edu/~wooland/count/count17.html 1. 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. 25,200 B. 3,628,800 C. 4,200 D. 604,800 The number of ways to choose 4 kids to get dimes is C(10, 4) = 210 The number of ways to choose 3 other kids to get dimes is C(6, 3) = 20 So, the number of ways to do both of these is 210×20 = 4,200 Solutions

  8. From The FUNDAMENTALIZER, Part 2 www.math.fsu.edu/~wooland/count/count17.html 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. 2. Suddenly, he feels too cheap to give away all his coins. Instead, he will choose either four kids to receive the dimes or three kids to receive the nickels. In how many ways may the coins be distributed among the children, assuming that no child will get more than one coin? The number of ways to choose 4 kids to get dimes is C(10, 4) = 210 If he doesn’t do that, the number of ways to choose 3 of the 10 kids to get dimes is C(10, 3) = 120 So, the number of ways to either choose 4 kids to get dimes or choose 3 kids to get nickels is 210 + 120 = 330 Solutions

  9. There are nine servers and six bussers 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. 51 B. 540 C. 2160 D. 102 E. None of these Exercise

  10. There are nine servers and six bussers 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. 51 B. 540 C. 2160 D. 102 E. None of these We will choose 2 of the 9 servers and we will choose 2 of the 6 bussers. Because the 2 servers are getting different jobs (shop steward, secretary, respectively), the number of ways to choose them is P(9, 2) = 72. Likewise, the number of ways to choose 2 of the 6 bussers is them is P(6, 2) = 330. So, the number of ways to choose 2 servers and 2 bussers is 72×330 = 2160. Solution

  11. There are nine servers and six busserss 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 servers, or they will select a shop steward and a secretary from among the bussers. How many outcomes are possible? A. 51 B. 540 C. 2160 D. 102 E. None of these Exercise

  12. There are nine servers and six busserss 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 servers, or they will select a shop steward and a secretary from among the bussers. How many outcomes are possible? A. 51 B. 540 C. 2160 D. 102 E. None of these Refer to the previous solution for details. The difference this time is that we will choose either 2 servers or 2 bussers, so the number of outcomes is P(9, 2) + P(6, 2) = 72+30 = 102 Solutiion

  13. 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 girl’s 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 boy’s 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. 50 B. 600 C. 61 D. 900 EXERCISE

  14. 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 girl’s 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 boy’s 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. 50 B. 600 C. 61 D. 900 Every name is either a girl’s name or a boy’s name. The number of girl’s names is P(5, 2) = 20 Number number of boy’s names is P(6, 2) = 30. So the total number of names is 20 + 30 = 50. Solution

  15. 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 Exercise #5

  16. 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 If we will combine at least 4 of the 6 traits, then we will combine either 4 traits or 5 traits or 6 traits. The number of ways to combine exactly 4 traits is C(6, 4) = 15. The number of ways to combine exactly 5 traits is C(6, 5) = 6. The number of ways to combine exactly 6 traits is C(6, 5) = 1. So, the number of ways to combine either 4 or 5 or 6 traits is 15 + 6 + 1 = 22. Solution

  17. 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. 5112 B. 143 C. 2556 D. none of these Exercise #7

  18. 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. 5112 B. 143 C. 2556 D. none of these We must do two things: We must create a 3-part name for the cat scrubbing machine, and a different 3-part name for the dog poop scooper. For each of these chores, the number of different outcomes is 6×3×4 = 72. That is, there are 72 possible 3-syllable names for the cat scrubber, and 72 possible names for the dog poop scooper. However, because the dog poop scooper can’t have exactly the same name as the other product, one of the 72 possible names can’t be used (because it was already given to the cat scrubber). So, there are 71 names that are usable of the dog pooper scooper. So, the number of cats to choose a name for the cat scrubber and choose a different name for the poop scooper is 72×71 = 5112 Solution

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