Activity 3: Combinations

Transcription

1 MDM4U: Mathematics of Data Management, Grade 12, University Preparation Unit 5: Solving Problems Using Counting Techniques Activity 3: Combinations Combinations Assignment 1. Jessica is in a very big hurry. She needs 3 birthday cards and 2 wedding cards. She runs in to the corner store which has 12 birthday cards and 6 wedding cards. How many different combinations of cards might she leave with? 3300 The order she chooses them does not matter, so this is a combination. 3 birthday cards and 2 wedding cards We need to use the product rule since we are buying both birthday and wedding cards. 2. Kyle got 9 new gifts for his birthday. a. How many ways can he choose 4 of them to go to his dad s house? The order he chooses them does not matter, so this is a combination. Kyle needs to choose 4 from 9. = 126 b. How many ways can he choose 5 of them to go to his dad s house? = 126 Notice that you get the same answer. Kyle choosing 4 to take is the same as choosing 4 to leave which is why =. We will investigate this property further in the next activity. 3. There were 19 questions in your last assignment. If I want put 2 of them on your test, how many different combinations of questions could I choose? The order I chooses them does not matter, so this is a combination. 19 questions choose 2. = Note: The answers in the next question are quite large. Depending on your calculators capability it may give you scientific notation. At Governor Simcoe there is a robotics team, a programming team, and a drama club. The three groups are teaming up to put on a show for the grade 8 students. They need 12 people to volunteer. If there are 15 robotics team members, 7 programmers, and 11 thespians. How many ways can they choose a committee if a. there are no restrictions. 15 robotics team members, 7 programmers, and 11 thespians give us a total of 33 to choose the 12 people we need from. = b. there are an equal number from each group. If there are an equal number of each, that means we need 4 of each. So from the 15 robotics team members choose 4, AND from the 7 programmers choose 4, AND from the 11 thespians (actors/actresses) choose 4. = (1365)(35)(330)

2 = c. 6 robotics team members are chosen. 6 Robotics AND 6 others (from the 7 programmers, 11 thespians for a total of 18) = (5005)(18 564) = d. Mohamed must be included. Once Mohamed is chosen, there is one less person left to choose from (32) and one less person to choose (11). Note: Putting Mohamed on the committee is one choice, so you could think of it as M = e. Ryan must not be chosen. Solution 1: Direct reasoning You cannot choose Ryan, so there are only 32 (rather than 33) people to choose your 12 from. = Solution 2: Indirect Reasoning All - Ryan on the committee = Ryan not on the committee - = = All is the same as the answer to a). Ryan on the committee is the same as d), Mohamed on the committee. Notice that indirect reasoning is not as efficient as direct reasoning in this example. f. Jessica and Ryan cannot work together. Solution 1 - Indirect Reasoning All - Together = Apart - J R = = All is the same as a). To find the ways that they are together, put Jessica on the committee, put Ryan on the committee. You then need 10 more people (for your total of 12) and you have only 31 people left to choose from (two less). Solution 2 - Cases Just Jessica OR Just Ryan OR Neither J + R + = = Just Jessica - Put Jessica on the committee. You now have 32 people left, but if Jessica is on the committee, Ryan cannot be, so you are down to 31 people to choose from. You have only chosen 1 person (Jessica) so you still need 11 more for your committee of 12. Just Ryan is exactly the same.

3 Neither - If neither one of them can be chosen, you have 31 people left to make your committee from. You have yet to choose anyone, so you still need to choose 12. You will use the rule of sum to find the total of the 3 cases. g. at least one programmer is chosen. Solution 1 - Indirect Reasoning All - 0 programmers = At least one programmer = (1)( ) = If you want at least one programmer, the only type of committee that you do not want is the ones that do not have any programmers on it. So take the total, subtract the number of committees you don t want (the ones with 0 programmers) to find the number of committees that have at least one programmer. 0 Programmers - 7 programmers, you don t want any. This means you must choose your 12 committee members from the other 26 people. You do not have to include 7C0, as it is equal to one, but you may include it if it helps. Solution 2 - Cases 1 Programmer OR 2 Programmers OR 3 Programmers OR... OR 7 Programmers = (7)( ) + (21)( ) + (35)( ) + (35)( ) + (21)( ) + (7)( ) + (1)(65 780) = This solution is significantly longer, but it will work! At least 1 programmer implies 1 or more. There are only 7 programmers, so the most you can have is 7. Don t forget that if you have 1 programmer, you must choose 11 others for a total of 12. If you choose 2 programmers you need 10 others for a total of 12. Notice in each case the total of your n values add to the total of 33 people and each of your r values add to 12, the total you need to choose. Notice that 7C2 = 7C5, and 7C3 = 7C4. Choosing 2 programmers to be on the committee is the same as choosing 5 programmers to not be on the committee. h. at least 2 robotics members are chosen. Solution 1 - Indirect Reasoning All - 0 robotics - 1 robotics = At least 2 robotics = (1)(18 564) - (15)(31 824) = If you want at least 2 robotics team members, you do not want ones have no robotics team members and you also don t want the ones with a single robotics team member on it. So take the total, subtract the number of committees you don t want (the ones with 0 robotics team members and the ones with 1 robotics team member) to find the number of committees that have at least two robotics team members. 0 Robotics - 15 robotics members, you don t want any. This means you must choose your 12 committee members from the other 18 people.

4 1 Robotics - 15 robotics members chose 1 AND 11 others from the other 18. Sometimes students get confused part way through the question. As they are calculating the number of committees with one robotics team member, they think wait why am I doing this, I don t want that. Remember with in direct reasoning you are finding the opposite (complement) and subtracting it from the total. Solution 2 - Cases 2 Robotics OR 3 Robotics OR 4 Robotics OR 5 Robotics OR...OR 12 Robotics = (105)(43758) + (455)(48620) + (1365)(43758) + (3003)(31824) + (5005)(18564) + (6435)(8568) + (6435)(3060) + (5005)(816) + (3003)(135) + (1368)(18) + (455)(1) = This is again entirely too time consuming. In the last example there were only 7 programmers so when we found at least 1, we stopped at 7 even though our committee needed 12. This time there are 15 Robotics team members and so we could have a group made up ENTIRELY of robotics team members. For at least 2 we have to consider having 2 through 12 robotics team members. This is not effective at all which is why it is important that you understand indirect reasoning. Although this calculation is time consuming, it can be shortened if you realize that 15C3 = 15C12, 15C5 = 15C10... i. there must be between 2 and 4 robotics members chosen 2 Robotics OR 3 Robotics OR 4 Robotics = (105)(43758) + (455)(48620) + (1365)(43758) = In this last example the number of cases you need to look at is quite reasonable. Again don t forget that you need to always choose a total of 12 in each case. Hopefully you realize that there is a pattern to these questions. Often the most difficult task you have is to determine if you are looking at a permutation or a combination. 5. a. There are 13 forwards on the Niagara Ice Dogs. The coach needs to make 4 lines consisting of 3 forwards. He thinks that he will be able to try every possible combination of forwards. Find the number of combinations of forwards the coach can make and determine if this is feasible in a 62 game season. There are a total of combinations which is not feasible. If we assume a 62 game season at 60 minutes in length played at even strength for the entire time, there are only 3720 minutes in the season. First line AND Second line AND Third line AND Last line = (286)(120)(35)(4) = The coach starts with 13 players and chooses his first line combination of 3. He is then left with 10 players from which to use his next line of 3. There are then 7 players for the third line and 4 for the last line. Notice that one player is not chosen. That is fine. It just means the coach still has 4 choices for his last line rather than just 1 like the example in the content.

5 We are again using the product rule to find the total combinations. b. There are 6 starters lined up on the blue line for the singing of O Canada. How many ways can they line up? 6! = 720 This is a permutation. There are 6 objects and you need to arrange all 6. Remember that there are two other ways to solve this. See activity 2 if you need to see the other ways to get the answer of 720. c. There are a total of 22 players on the Ice Dog roster. How many ways can the coach choose 7 of them to go to an elementary school for their anti-bullying campaign? = players, choose 7. It s as simple as that. Students tend to over complicate questions when they think that they should be getting harder. Short, easy questions can pop up as a part of a harder question at any time. d. There are a total of 22 players on the Ice Dog roster. How many ways can he send 7 to one school, 4 to another, 8 to another, and 3 to the last school? = (170544)(1365)(165)(1) = You may leave you answers in scientific notation, but if you are planning to take more courses in mathematics, make sure you remember how to convert scientific notation to standard notation. This uses the same thinking as a), but this time since the coach is using all of the students, you could drop off choosing the last school as the coach is just sending the remaining 3 players. e. There are a total of 22 players on the Ice Dog roster. How many ways can the coach choose a captain, a first assistant, and a second assistant? 22P3 or = 9240 This is again a permutation. He is choosing specific roles for the players, so order is important. Another acceptable answer is to use both permutations and combinations. In the question there are three specific roles - Captain, 1st Assistant, and 2nd Assistant. If you watch or play hockey you may think that 1st Assistant and 2nd Assistant are not specific roles, they are just the assistants. If you thought of the question that way the answer is: Choose Captain AND Choose 2 Assistants = 22P1 21C2 = 4620 Notice that there are less choices when there are less distinct roles because it no longer matters if you are the first assistant or the second assistant. 6. A Final Exam in English is broken into 3 parts. Part A consists of 3 Written Passages, Part B consists of 8 Short Answer, and Part C consists of 4 Essay Questions. How many different combinations of questions are possible if you must answer nine questions and: a. there are no restrictions = = 15 questions in total. 15 choose 9 to answer. b. you must answer 2 from Part A, 6 from Part B, and 1 from Part C? 2A AND 6B AND 1C = (3)(28)(4) = 336 c. you must answer at least 1 Essay question This questions uses the exact same thinking as 4g.

6 Solution 1: Indirect Reasoning All - 0 Essay = At least one Essay = = 4950 This uses the exact same thinking as 4g. Solution 2: Cases 1 Essay AND 8 others OR 2 Essay AND 7 others OR 3 Essay AND 6 others OR 4 Essay AND 5 others = (4)(165) + (6)(330) + (4)(462) + (1)(462) = 4950 d. you must answer at least 2 of the short answer questions. This uses the exact same thinking as 4h, however you may encounter a problem. If you answer 0 short answer questions, you still need to answer 9, but there are only 7 left. This can t happen, so there are no combinations with 0 short answer questions. The same is true if you only answer 1. Solution 1: Indirect Reasoning All - 0 Short Answer - 1 Short Answer = At least two Short Answer = = 5005 When a student gets stuck on the indirect reasoning above because it won t work, they will then try direct reasoning. It is important that you are comfortable with both solutions. Solution 2: Cases 2 Short Answer OR 3 Short Answer OR 4 Short Answer OR 5 Short Answer...OR 8 Short Answer = (28)(1) + (56)(7) + (70)(21) + (56)(35) + (28)(35) + (8)(21) + (1)(7) = 5005 Solution 3: Logic! Above we discussed that you can t answer 9 questions without answering at least 2 from the short answer section as there are only 7 other questions. Using this logic, all possible combinations (15C9) have at least 2 short answer questions in them. 7. In a deck of 52 cards, how many ways are there to: (You may leave your answers in the form or ncr) a. deal 5 cards. There are a total of 52 cards in a deck. Those 52 cards are spit so that half (26) are red and half are black. The 52 cards are also separated into 4 suits of 13 cards each. Those suits are diamonds and hearts (red) and spades and clubs (black). Within each suits the 13 cards are as follows: Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King. The Jack King, and Queen are commonly referred to as face cards. A common application of counting problems is different types of games, including casino games. If you are not familiar with a deck of cards you should research the different suits as you will encounter cards questions in your next unit of study on probability.

7 b. deal 5 cards to one player, then deal 5 to a second. First hand AND Second hand Just like in the Ice Dog question, once you deal 5 to the first player there are less cards to choose from for the second player. c. deal 3 hearts and 2 diamonds. 3 hearts AND 2 diamonds There are 13 cards in each suit. So you have 13 cards to choose the 3 hearts from and 13 cards to choose your 2 diamonds from. d. deal exactly 3 face cards if you are dealing a total of 5. 3 Face cards AND 2 others There are 3 face cards (Jack, Queen, King) in each of the 4 suits, for a total of 12. A common mistake is to forget that you still need to choose the other 2 cards. If 12 cards are face cards then the complement (the cards that are not face cards) are the other 40. e. deal only red cards if you are being dealt five cards. There are 26 red cards in the deck. If you only want red cards then you are choosing your 5 from only those 26. Close Window