2. PERMUTATIONS AND COMBINATIONS

3. BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED FACTORIAL.

4. A FACTORIAL is a counting method that uses consecutive whole numbers as factors. The factorial symbol is ! Examples 5! = 5x4x3x2x1 = 120 7! = 7x6x5x4x3x2x1 = 5040

5. First, we’ll do some permutation problems. Permutations are “arrangements”.

6. Permutations In the context of counting problems, arrangements are often called permutations; the number of permutations of n things taken r at a time is denoted n P r. Applying the fundamental counting principle to arrangements of this type gives n P r = n(n – 1)(n – 2)…[n – (r – 1)].

7. Factorial Formula for Permutations The number of permutations, or arrangements, of n distinct things taken r at a time, where r n, can be calculated as

8. Example: Permutations Evaluate each permutation. a) 5 P 3 b) 6 P 6 Solution

9. Example: IDs How many ways can you select two letters followed by three digits for an ID if repeats are not allowed? Solution There are two parts: 1. Determine the set of two letters. 2. Determine the set of three digits. Part 1 Part 2

10. Example: Building Numbers From a Set of Digits How many four-digit numbers can be written using the numbers from the set {1, 3, 5, 7, 9} if repetitions are not allowed? Solution Repetitions are not allowed and order is important, so we use permutations:

11. Let’s do a permutation problem. How many different arrangements are there for 3 books on a shelf? Books A,B, and C can be arranged in these ways: ABC ACB BAC BCA CAB CBA Six arrangements or 3! = 3x2x1 = 6

12. In a permutation, the order of the books is important. Each different permutation is a different arrangement. The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.

13. You try this one: 1. How many ways can 4 books be arranged on a shelf? 4! or 4x3x2x1 or 24 arrangements Here are the 24 different arrangements: ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBAC DBCA DCAB DCBA

14. Now we’re going to do 3 books on a shelf again, but this time we’re going to choose them from a group of 8 books. We’re going to have a lot more possibilities this time, because there are many groups of 3 books to be chosen from the total 8, and there are several different arrangements for each group of 3.

15. If we were looking for different arrangements for all 8 books, then we would do 8! But we only want the different arrangements for groups of 3 out of 8, so we’ll do a partial factorial, 8x7x6 =336

16. Try these: 1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there? 10x9x8x7x6 or 30240

17. 2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there? 7x6x5x4 or 840

18. Combinations

19. In the context of counting problems, subsets, where order of elements makes no difference, are often called combinations; the number of combinations of n things taken r at a time is denoted n C r.

20. Factorial Formula for Combinations The number of combinations, or subsets, of n distinct things taken r at a time, where r n, can be calculated as Note:

21. Example: Combinations Evaluate each combination. a) 5 C 3 b) 6 C 6 Solution

22. Now, we’ll do some combination problems. Combinations are “selections”.

23. There are some problems where the order of the items is NOT important. These are called combinations. You are just making selections, not making different arrangements.

24. Example: A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed? Let’s call the 5 people A,B,C,D,and E. Suppose the selected committee consists of students E, C, and A. If you re- arrange the names to C, A, and E, it’s still the same group of people. This is why the order is not important.

25. Because we’re not going to use all the possible combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees. Therefore instead of using only 5x4x3, to get the fewer committees, we must divide. 5x4x3 3x2x1 (Always divide by the factorial of the number of digits on top of the fraction.) Answer: 10 committees

26. Now, you try. 1. How many possible committees of 2 people can be selected from a group of 8? 8x7 2x1 or 28 possible committees

27. 2. How many committees of 4 students could be formed from a group of 12 people? 12x11x10x9 4x3x2x1 or 495 possible committees

28. Example: Finding the Number of Subsets Find the number of different subsets of size 3 in the set {m, a, t, h, r, o, c, k, s}. Solution A subset of size 3 must have 3 distinct elements, so repetitions are not allowed. Order is not important.

29. Example: Finding the Number of Poker Hands A common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible? Solution Repetitions are not allowed and order is not important.

30. Guidelines on Which Method to Use PermutationsCombinations Number of ways of selecting r items out of n items Repetitions are not allowed Order is important.Order is not important. Arrangements of n items taken r at a time Subsets of n items taken r at a time n P r = n!/(n – r)! n C r = n!/[ r!(n – r)!] Clue words: arrangement, schedule, order Clue words: group, sample, selection

31. Permutations and Combinations Evaluate each problem. c) 6 P 6 a) 5 P 3 b) 5 C 3 d) 6 C 6 543543 60 10 720 1

32. Permutations and Combinations How many ways can you select two letters followed by three digits for an ID if repeats are not allowed? Two parts: 2. Determine the set of three digits.1. Determine the set of two letters. 26 P 210 P 3 26  25 650 10  9  8 720 650  720 468,000

33. Permutations and Combinations A common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible? Hint: Repetitions are not allowed and order is not important. 52 C 5 2,598,960 5-card hands

34. Permutations and Combinations Find the number of different subsets of size 3 in the set: {m, a, t, h, r, o, c, k, s}. Find the number of arrangements of size 3 in the set: {m, a, t, h, r, o, c, k, s}. 9C39C3 84 Different subsets 9P39P3 987987 504arrangements

35. 10.3 – Using Permutations and Combinations Guidelines on Which Method to Use

36. CIRCULAR PERMUTATIONS When items are in a circular format, to find the number of different arrangements, divide: n! / n

37. Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there? 6!  6 = 120

38. Fundamental Counting Principle For a college interview, Robert has to choose what to wear from the following: 4 slacks, 3 shirts, 2 shoes and 5 ties. How many possible outfits does he have to choose from? 4*3*2*5 = 120 outfits

39. 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?

40. Permutations 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? Practice:

41. 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?

42. Permutations 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? Answer:

43. 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?

44. 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? Answer:

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

46. 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? Answer:

47. 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?

48. Combinations 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? Answer: Center: Forwards:Guards: Thus, the number of ways to select the starting line up is 2*10*6 = 120.

49. How many ways can a student government select a president, vice president, secretary, and treasurer from a group of 6 people? This is the equivalent of selecting and arranging 4 items from 6. = 6 5 4 3 = 360 Divide out common factors. There are 360 ways to select the 4 people. Substitute 6 for n and 4 for r in

50. How many ways can a stylist arrange 5 of 8 vases from left to right in a store display? Divide out common factors. = 8 7 6 5 4 = 6720 There are 6720 ways that the vases can be arranged.

51. Awards are given out at a costume party. How many ways can “most creative,” “silliest,” and “best” costume be awarded to 8 contestants if no one gets more than one award? = 8 7 6 = 336 There are 336 ways to arrange the awards.

52. How many ways can a 2-digit number be formed by using only the digits 5–9 and by each digit being used only once? = 5 4 = 20 There are 20 ways for the numbers to be formed.

53. 1. Six different books will be displayed in the library window. How many different arrangements are there? 2. The code for a lock consists of 5 digits. The last number cannot be 0 or 1. How many different codes are possible? 80,000 720 3. The three best essays in a contest will receive gold, silver, and bronze stars. There are 10 essays. In how many ways can the prizes be awarded? 4. In a talent show, the top 3 performers of 15 will advance to the next round. In how many ways can this be done? 455 720

54. Tournament Problems

55. An amusement park has 27 different rides. If you have 21 ride tickets, how many different combinations of rides can you take?

56. Answer: 296,010

57. Pop’s Pizza offers 4 types of meat and 3 types of cheese. In how many ways could a pizza with two meats, different or double of the same meat, and one cheese be ordered?

58. Answer: 30

59. Pop’s Pizza offers 4 types of meat and 3 types of cheese. In how many ways could a pizza with two meats, different or double of the same meat, and one cheese be ordered?

60. Answer: 30