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Permutations and Combinations

Permutations and Combinations. Key Vocabulary: Permutation Combination Fundamental Counting Principle Factorial. Objectives:. apply fundamental counting principle compute permutations compute combinations distinguish permutations vs combinations.

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Permutations and Combinations

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  1. Permutations andCombinations Key Vocabulary: Permutation Combination Fundamental Counting Principle Factorial

  2. Objectives: • apply fundamental counting principle • compute permutations • compute combinations • distinguish permutations vs combinations

  3. Fundamental Counting Principle Lets start with a simple example. A student is to roll a die and flip a coin. How many possible outcomes will there be? 1H 2H 3H 4H 5H 6H 1T 2T 3T 4T 5T 6T 12 outcomes 6*2 = 12 outcomes

  4. Example: For dinner you have the following choices: ENTREES MAINS chicken soup salad prawns hamburger DESSERTS How many different combinations of meals could you make? We'll build a tree diagram to show all of the choices.  icecream

  5. We ended up with 12 possibilities ice cream  chicken ice cream prawns  hamburger soup ice cream  ice cream salad  chicken ice cream prawns  hamburger ice cream  Notice the number of choices at each branch 2choices 3choices 2choices soup, chicken, ice cream soup, chicken,  2  3  2 = 12 soup, prawns, ice cream soup, prawns,  soup, hamburger, ice cream soup, hamburger,  salad, chicken, ice cream salad, chicken,  salad, prawns, ice cream salad, prawns,  Now to get all possible choices we follow each path. salad, hamburger, ice cream salad, hamburger, 

  6. Multiplication Principle of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in different ways. pqr If we have 6 different shirts, 4 different pants, 5 different pairs of socks and 3 different pairs of shoes, how many different outfits could we wear? 6  4  5  3 = 360

  7. Permutations A Permutation is an arrangement of items in a particular order. Notice, ORDER MATTERS! To find the number of Permutations of n items, we can use the Fundamental Counting Principle or factorial notation.

  8. Permutations The number of ways to arrange the letters ABC: ____ ____ ____ 3 ____ ____ Number of choices for first blank? 3 2 ___ Number of choices for second blank? 321 Number of choices for third blank? 3*2*1 = 6 3! = 3*2*1 = 6 ABC ACB BAC BCA CAB CBA

  9. A permutation is an ordered arrangement of r objects chosen from n objects. For combinations order does not matter but for permutations it does. There are three types of permutations.The first is distinctwithrepetition. This means there are n distinct objects but in choosing r of them you can repeat an object. this means different Let's look at a 3 combination lock with numbers 0 through 9 There are 10 choices for the first number There are 10 choices for the second number and you can repeat the first number There are 10 choices for the third number and you can repeat By the multiplication principle there are 10  10  10 = 1000 choices

  10. This can be generalized as: Permutations: Distinct Objects with Repetition The number of ordered arrangements of r objects chosen from n objects, in which the n objects are distinct and repetition is allowed, is nr What if the lock had four choices for numbers instead of three? 104 = 10 000 choices

  11. 1st 2nd 3rd 4th The second type of permutation is distinct, without repetition. Let's say four people have a race. Let's look at the possibilities of how they could place. Once a person has been listed in a place, you can't use that person again (no repetition). Based on the multiplication principle: 4  3  2 1 = 24 choices First place would be choosing someone from among 4 people. Now there are only 3 to choose from for second place. Now there are only 2 to choose from for third place. Only one possibility for fourth place.

  12. nPr , means the number of ordered arrangements of r objects chosen from n distinct objects and repetition is not allowed. In the last example: If you have 10 people racing and only 1st, 2nd and 3rd place how many possible outcomes are there? 0! = 1

  13. Permutations To find the number of Permutations of n items chosen r at a time, you can use the formula

  14. Answer Now Permutations Practice: A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

  15. Permutations Practice: A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

  16. Answer Now Permutations Practice: From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled?

  17. Permutations Practice: From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled?

  18. Combinations A Combination is an arrangement of items in which order does not matter. ORDER DOES NOT MATTER! Since the order does not matter in combinations, there are fewer combinations than permutations.  The combinations are a "subset" of the permutations.

  19. Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula

  20. Combinations To find the number of Combinations of n items chosen r at a time, you can use the formula

  21. A combination is an arrangement of r objects chosen from n objects regardless of order. nCr, means the number combinations of r objects chosen from n distinct objects and repetition is not allowed. Order doesn't matter here so the combination 1, 2, 3 is not different than 3, 2, 1 because they both contain the same numbers.

  22. You need 2 people on your committee and you have 5 to choose from. You can see that this is without repetition because you can only choose a person once, and order doesn’t matter. You need 2 committee members but it doesn't matter who is chosen first. How many combinations are there?

  23. Answer Now Combinations Practice: To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?

  24. Combinations To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? Practice:

  25. Answer Now Combinations Practice: A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?

  26. Combinations A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? Practice:

  27. Answer Now Combinations Practice: A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards?

  28. Combinations Practice: A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Guards: Forwards: Center: Thus, the number of ways to select the starting line up is 2*10*6 = 120.

  29. Review/Summary • Permutations vs. Combinations When is it a permutation? When is it a combination?

  30. Permutation A permutation is an arrangement in which order matters. A B C differs from B C A

  31. How Many Permutations? Consider four objects {A,B,C,D} There are 4 choices for the first slot. There are 3 choices for the second slot. There are 2 choices for the third slot. There is 1 choice for the last slot.

  32. 4 x 3 x 2 x 1 = 24 Permutations ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBAC DBCA DCAB DCBA

  33. Generalization There are 4! ways to arrange 4 items. There are n! ways to arrange n items.

  34. Permutation Example Selecting 3 items out of a set of 5 We have 5 choices for the first item. We have 4 choices for the second item. We have 3 choices for the third item. 5 x 4 x 3 = 60 Permutations

  35. Permutation Formula

  36. Combinations Combinations are arrangements in which order does NOT matter. A, B, C is the same as B, C, A

  37. Evaluating In how many ways may 3 items be selected from a set of 5 without regard to order?

  38. We already know that there 60permutations of these items. For each set of three, there are 3! or 6 arrangements. A B C A C B B A C B C A C A B C B A All of these are really the same.

  39. Our actual answer is 10. Consider the set {A,B,C,D,E} These are the combinations. A,B,C A,B,D A,B,E A,C,D A,C,E A,D,E B,C,D B,C,E B,D,E C,D,E

  40. Combination Formula

  41. A Challenging Example. Have a go.  Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? We could have even numbers with 4, 5 or 6 digits This Gives 4 possibilities to work with: PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 PART B: 4, 5 or 6 EVEN digits beginning with a 5 PART C: 5 or 6 EVEN digits beginning with a 2 PART D: 5 or 6 EVEN digits beginning with a 1 or 3

  42. A Challenging Example. Have a go.  Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 2 4 3 2 + 2 4 3 2 2 + 2 4 3 2 1 2 This gives a total of 240

  43. A Challenging Example. Have a go.  Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? PART B: 4, 5 or 6 EVEN digits beginning with a 5 1 4 3 3 + 1 4 3 2 3 + 1 4 3 2 1 3 This gives a total of 180

  44. A Challenging Example. Have a go.  Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? PART C: 5 or 6 EVEN digits beginning with a 2 1 4 3 2 2 + 1 4 3 2 1 2 This gives a total of 96

  45. A Challenging Example. Have a go.  Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? PART D: 5 or 6 EVEN digits beginning with a 1 or 3 2 4 3 2 3 + 2 4 3 2 1 3 This gives a total of 288

  46. A Challenging Example. Have a go.  Permutation: Order Matters How many even numbers greater than 4000 can be formed using some or all of the digits 1, 2, 3, 4, 5, 6 if each digit must feature no more than once in a number? We could have even numbers with 4, 5 or 6 digits This Gives 4 possibilities to work with: PART A: 4, 5 or 6 EVEN digits beginning with a 4 OR 6 = 240 PART B: 4, 5 or 6 EVEN digits beginning with a 5 = 180 PART C: 5 or 6 EVEN digits beginning with a 2 = 96 PART D: 5 or 6 EVEN digits beginning with a 1 or 3 =288 Number of possible even numbers greater than 4000 = 804

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