Math 1300: Section 7- 4 Permutations and Combinations
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The document discusses permutations and combinations. It provides definitions and examples of factorials, permutations of n objects taken r at a time, and combinations of n objects taken r at a time. It includes theorems on calculating the number of permutations and combinations. Examples are provided to demonstrate calculating permutations and combinations for different values of n and r.
Math 1300: Section 7- 4 Permutations and Combinations
- 1. Factorials Permutations Combinations Math 1300 Finite Mathematics Section 7-4: Permutations and Combinations Jason Aubrey Department of Mathematics University of Missouri university-logo Jason Aubrey Math 1300 Finite Mathematics
- 2. Factorials Permutations Combinations Problem 1: Consider the set {p, e, n}. How many two-letter "words" (including nonsense words) can be formed from the members of this set, if two different letters have to be used? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 3. Factorials Permutations Combinations Problem 1: Consider the set {p, e, n}. How many two-letter "words" (including nonsense words) can be formed from the members of this set, if two different letters have to be used? We list all possibilities: pe, pn, en, ep, np, ne, a total of 6. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 4. Factorials Permutations Combinations Problem 1: Consider the set {p, e, n}. How many two-letter "words" (including nonsense words) can be formed from the members of this set, if two different letters have to be used? We list all possibilities: pe, pn, en, ep, np, ne, a total of 6. Problem 2: Now consider the set consisting of three males: {Paul, Ed, Nick }. For simplicity, denote the set by {p, e, n}. How many two-man crews can be selected from this set? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 5. Factorials Permutations Combinations Problem 1: Consider the set {p, e, n}. How many two-letter "words" (including nonsense words) can be formed from the members of this set, if two different letters have to be used? We list all possibilities: pe, pn, en, ep, np, ne, a total of 6. Problem 2: Now consider the set consisting of three males: {Paul, Ed, Nick }. For simplicity, denote the set by {p, e, n}. How many two-man crews can be selected from this set? pe (Paul, Ed), pn (Paul, Nick) and en (Ed, Nick) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 6. Factorials Permutations Combinations Both problems involved counting the numbers of arrangements of the same set {p, e, n}, taken 2 elements at a time, without allowing repetition. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 7. Factorials Permutations Combinations Both problems involved counting the numbers of arrangements of the same set {p, e, n}, taken 2 elements at a time, without allowing repetition. However, in the first problem, the order of the arrangements mattered since pe and ep are two different "words". university-logo Jason Aubrey Math 1300 Finite Mathematics
- 8. Factorials Permutations Combinations Both problems involved counting the numbers of arrangements of the same set {p, e, n}, taken 2 elements at a time, without allowing repetition. However, in the first problem, the order of the arrangements mattered since pe and ep are two different "words". In the second problem, the order did not matter since pe and ep represented the same two-man crew. We counted this only once. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 9. Factorials Permutations Combinations Both problems involved counting the numbers of arrangements of the same set {p, e, n}, taken 2 elements at a time, without allowing repetition. However, in the first problem, the order of the arrangements mattered since pe and ep are two different "words". In the second problem, the order did not matter since pe and ep represented the same two-man crew. We counted this only once. The first example was concerned with counting the number of permutations of 3 objects taken 2 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 10. Factorials Permutations Combinations Both problems involved counting the numbers of arrangements of the same set {p, e, n}, taken 2 elements at a time, without allowing repetition. However, in the first problem, the order of the arrangements mattered since pe and ep are two different "words". In the second problem, the order did not matter since pe and ep represented the same two-man crew. We counted this only once. The first example was concerned with counting the number of permutations of 3 objects taken 2 at a time. The second example was concerned with the number of combinations of 3 objects taken 2 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 11. Factorials Permutations Combinations Definition (Factorial) For n a natural number, n! = n(n − 1)(n − 2) · · · 2 · 1 0! = 1 n! = n · (n − 1)! Note: Many calculators have an n! key or its equivalent university-logo Jason Aubrey Math 1300 Finite Mathematics
- 12. Factorials Permutations Combinations Examples 3! = 3(2)(1) = 6 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 13. Factorials Permutations Combinations Examples 3! = 3(2)(1) = 6 4! = 4(3)(2)(1) = 12 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 14. Factorials Permutations Combinations Examples 3! = 3(2)(1) = 6 4! = 4(3)(2)(1) = 12 5! = 5(4)(3)(2)(1) = 120 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 15. Factorials Permutations Combinations Examples 3! = 3(2)(1) = 6 4! = 4(3)(2)(1) = 12 5! = 5(4)(3)(2)(1) = 120 6! = 6(5)(4)(3)(2)(1) = 720 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 16. Factorials Permutations Combinations Examples 3! = 3(2)(1) = 6 4! = 4(3)(2)(1) = 12 5! = 5(4)(3)(2)(1) = 120 6! = 6(5)(4)(3)(2)(1) = 720 7! = 7(6)(5)(4)(3)(2)(1) = 5040 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 17. Factorials Permutations Combinations Definition (Permutation of a Set of Objects) A permutation of a set of distinct objects is an arrangement of the objects in a specific order without repetition. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 18. Factorials Permutations Combinations Theorem (Number of Permutations of n Objects) The number of permutations of n distinct objects without repetition, denoted by Pn,n is Pn,n = n(n − 1) · · · 2 · 1 = n! university-logo Jason Aubrey Math 1300 Finite Mathematics
- 19. Factorials Permutations Combinations Example: Now suppose that the director of the art gallery decides to use only 2 of the 4 available paintings, and they will be arranged on the wall from left to right. We are now talking about a particular arrangement of 2 paintings out of the 4, which is called a permutation of 4 objects taken 2 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 20. Factorials Permutations Combinations Example: Now suppose that the director of the art gallery decides to use only 2 of the 4 available paintings, and they will be arranged on the wall from left to right. We are now talking about a particular arrangement of 2 paintings out of the 4, which is called a permutation of 4 objects taken 2 at a time. × university-logo Jason Aubrey Math 1300 Finite Mathematics
- 21. Factorials Permutations Combinations Example: Now suppose that the director of the art gallery decides to use only 2 of the 4 available paintings, and they will be arranged on the wall from left to right. We are now talking about a particular arrangement of 2 paintings out of the 4, which is called a permutation of 4 objects taken 2 at a time. 4× university-logo Jason Aubrey Math 1300 Finite Mathematics
- 22. Factorials Permutations Combinations Example: Now suppose that the director of the art gallery decides to use only 2 of the 4 available paintings, and they will be arranged on the wall from left to right. We are now talking about a particular arrangement of 2 paintings out of the 4, which is called a permutation of 4 objects taken 2 at a time. 4×3 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 23. Factorials Permutations Combinations Example: Now suppose that the director of the art gallery decides to use only 2 of the 4 available paintings, and they will be arranged on the wall from left to right. We are now talking about a particular arrangement of 2 paintings out of the 4, which is called a permutation of 4 objects taken 2 at a time. 4 × 3 = 12 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 24. Factorials Permutations Combinations Definition (Permutation of n Objects Taken r at a Time) A permutation of a set of n distinct objects taken r at a time without repetition is an arrangement of r of the n objects in a specific order. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 25. Factorials Permutations Combinations Theorem (Number of Permutations of n Objects Taken r at a Time) The number of permutations of n distinct objects taken r at a time without repetition is given by Pn,r = n(n − 1)(n − 2) · · · (n − r + 1) or n! Pn,r = 0≤r ≤n (n − r )! n! n! Note: Pn,n = (n−n)! = 0! = n! permutations of n objects taken n at a time. Note: In place of Pn,r the symbol P(n, r ) is often used. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 26. Factorials Permutations Combinations Example: Find P(5, 3) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 27. Factorials Permutations Combinations Example: Find P(5, 3) P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 28. Factorials Permutations Combinations Example: Find P(5, 3) P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60. This means there are 60 permutations of 5 items taken 3 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 29. Factorials Permutations Combinations Example: Find P(5, 3) P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60. This means there are 60 permutations of 5 items taken 3 at a time. Example: Find P(5, 5), the number of permutations of 5 objects taken 5 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 30. Factorials Permutations Combinations Example: Find P(5, 3) P(5, 3) = (5)(5 − 1)(5 − 2) = (5)(4)(3) = 60. This means there are 60 permutations of 5 items taken 3 at a time. Example: Find P(5, 5), the number of permutations of 5 objects taken 5 at a time. P(5, 5) = 5(4)(3)(2)(1) = 120. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 31. Factorials Permutations Combinations Example: A park bench can seat 3 people. How many seating arrangements are possible if 3 people out of a group of 5 sit down? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 32. Factorials Permutations Combinations Example: A park bench can seat 3 people. How many seating arrangements are possible if 3 people out of a group of 5 sit down? P(5, 3) = (5)(4)(3) = 60. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 33. Factorials Permutations Combinations Example: A bookshelf has space for exactly 5 books. How many different ways can 5 books be arranged on this bookshelf? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 34. Factorials Permutations Combinations Example: A bookshelf has space for exactly 5 books. How many different ways can 5 books be arranged on this bookshelf? P(5, 5) = 5(4)(3)(2)(1) = 120 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 35. Factorials Permutations Combinations Definition (Combination of n Objects Taken r at a Time) A combination of a set of n distinct objects taken r at a time without repetition is an r -element subset of the set of n objects. The arrangement of the elements in the subset does not matter. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 36. Factorials Permutations Combinations Theorem (Number of Combinations of n Objects Taken r at a Time) The number of combinations of n distinct objects taken r at a time without repetition is given by: n Cn,r = r Pn,r = r! n! = 0≤r ≤n r !(n − r )! n Note: In place of the symbols Cn,r and , the symbols r university-logo C(n, r ) is often used. Jason Aubrey Math 1300 Finite Mathematics
- 37. Factorials Permutations Combinations Examples: P(8,5) 8(7)(6)(5)(4) C(8, 5) = 5! = 5(4)(3)(2)(1) = 56 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 38. Factorials Permutations Combinations Examples: P(8,5) 8(7)(6)(5)(4) C(8, 5) = 5! = 5(4)(3)(2)(1) = 56 P(8,8) 8(7)(6)(5)(4)(3)(2)(1) C(8, 8) = 8! = 8(7)(6)(5)(4)(3)(2)(1) =1 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 39. Factorials Permutations Combinations Example: In how many ways can you choose 5 out of 10 friends to invite to a dinner party? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 40. Factorials Permutations Combinations Example: In how many ways can you choose 5 out of 10 friends to invite to a dinner party? Does the order of selection matter? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 41. Factorials Permutations Combinations Example: In how many ways can you choose 5 out of 10 friends to invite to a dinner party? Does the order of selection matter? No! So we use combinations. . . university-logo Jason Aubrey Math 1300 Finite Mathematics
- 42. Factorials Permutations Combinations Example: In how many ways can you choose 5 out of 10 friends to invite to a dinner party? Does the order of selection matter? No! So we use combinations. . . P(10, 5) 10(9)(8)(7)(6) C(10, 5) = = = 252 5! 5(4)(3)(2)(1) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 43. Factorials Permutations Combinations Example: How many 5-card poker hands will have 3 aces and 2 kings? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 44. Factorials Permutations Combinations Example: How many 5-card poker hands will have 3 aces and 2 kings? The solution involves both the multiplication principle and combinations. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 45. Factorials Permutations Combinations Example: How many 5-card poker hands will have 3 aces and 2 kings? The solution involves both the multiplication principle and combinations. O1 : Choose 3 aces out of 4 N1 : C4,3 O2 : Choose 2 kings out of 4 N2 : C4,2 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 46. Factorials Permutations Combinations Example: How many 5-card poker hands will have 3 aces and 2 kings? The solution involves both the multiplication principle and combinations. O1 : Choose 3 aces out of 4 N1 : C4,3 O2 : Choose 2 kings out of 4 N2 : C4,2 Using the multiplication principle, we have: number of hands = C4,3 C4,2 4! 4! = 3!(4 − 3)! 2!(4 − 2)! = 4 · 6 = 24 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 47. Factorials Permutations Combinations Example: A computer store receives a shipment of 24 laser printers, including 5 that are defective. Three of these printers are selected to be displayed in the store. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 48. Factorials Permutations Combinations Example: A computer store receives a shipment of 24 laser printers, including 5 that are defective. Three of these printers are selected to be displayed in the store. (a) How many selections can be made? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 49. Factorials Permutations Combinations Example: A computer store receives a shipment of 24 laser printers, including 5 that are defective. Three of these printers are selected to be displayed in the store. (a) How many selections can be made? P(24, 3) C(24, 3) = 3! university-logo Jason Aubrey Math 1300 Finite Mathematics
- 50. Factorials Permutations Combinations Example: A computer store receives a shipment of 24 laser printers, including 5 that are defective. Three of these printers are selected to be displayed in the store. (a) How many selections can be made? P(24, 3) C(24, 3) = 3! 24 · 23 · 22 = 3·2·1 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 51. Factorials Permutations Combinations Example: A computer store receives a shipment of 24 laser printers, including 5 that are defective. Three of these printers are selected to be displayed in the store. (a) How many selections can be made? P(24, 3) C(24, 3) = 3! 24 · 23 · 22 = 3·2·1 = 2024 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 52. Factorials Permutations Combinations (b) How many of these selections will contain no defective printers? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 53. Factorials Permutations Combinations (b) How many of these selections will contain no defective printers? P(19, 3) C(19, 3) = 3! university-logo Jason Aubrey Math 1300 Finite Mathematics
- 54. Factorials Permutations Combinations (b) How many of these selections will contain no defective printers? P(19, 3) C(19, 3) = 3! 19 · 18 · 17 = 3·2·1 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 55. Factorials Permutations Combinations (b) How many of these selections will contain no defective printers? P(19, 3) C(19, 3) = 3! 19 · 18 · 17 = 3·2·1 = 969 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 56. Factorials Permutations Combinations Example: Suppose a certain bank has 4 branches in Columbia, 12 branches in St. Louis, and 10 branches in Kansas City. The bank must close 6 of these branches. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 57. Factorials Permutations Combinations Example: Suppose a certain bank has 4 branches in Columbia, 12 branches in St. Louis, and 10 branches in Kansas City. The bank must close 6 of these branches. (a) In how many ways can this be done? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 58. Factorials Permutations Combinations Example: Suppose a certain bank has 4 branches in Columbia, 12 branches in St. Louis, and 10 branches in Kansas City. The bank must close 6 of these branches. (a) In how many ways can this be done? C(26, 6) = 230, 230 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 59. Factorials Permutations Combinations Example: Suppose a certain bank has 4 branches in Columbia, 12 branches in St. Louis, and 10 branches in Kansas City. The bank must close 6 of these branches. (a) In how many ways can this be done? C(26, 6) = 230, 230 (b) The bank decides to close 2 branches in Columbia, 3 branches in St. Louis, and 1 branch in Kansas City. In how many ways can this be done? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 60. Factorials Permutations Combinations Example: Suppose a certain bank has 4 branches in Columbia, 12 branches in St. Louis, and 10 branches in Kansas City. The bank must close 6 of these branches. (a) In how many ways can this be done? C(26, 6) = 230, 230 (b) The bank decides to close 2 branches in Columbia, 3 branches in St. Louis, and 1 branch in Kansas City. In how many ways can this be done? O1 : pick Columbia branches N1 : C(4, 2) O2 : pick STL branches N2 : C(12, 3) O3 : pick KC branches N3 : C(10, 1) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 61. Factorials Permutations Combinations O1 : pick Columbia branches N1 : C(4, 2) O2 : pick STL branches N2 : C(12, 3) O3 : pick KC branches N3 : C(10, 1) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 62. Factorials Permutations Combinations O1 : pick Columbia branches N1 : C(4, 2) O2 : pick STL branches N2 : C(12, 3) O3 : pick KC branches N3 : C(10, 1) By the multiplication principle Number of ways = C(4, 2)C(12, 3)C(10, 1) = 6 · 220 · 10 = 13, 200 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 63. Factorials Permutations Combinations Example (FS09 Final Exam) During the semester, a professor wrote various problems to possibly include on the final exam in a math course. The course covered 4 chapters of a textbook. She wrote 2 problems from Chapter 1, 3 problems from Chapter 2, 5 problems from Chapter 3, and 4 problems from Chapter 4. She decides to put two questions from each chapter on the final exam. university-logo Jason Aubrey Math 1300 Finite Mathematics
- 64. Factorials Permutations Combinations Example (FS09 Final Exam) During the semester, a professor wrote various problems to possibly include on the final exam in a math course. The course covered 4 chapters of a textbook. She wrote 2 problems from Chapter 1, 3 problems from Chapter 2, 5 problems from Chapter 3, and 4 problems from Chapter 4. She decides to put two questions from each chapter on the final exam. (a) In how many different ways can she choose two questions from each chapter? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 65. Factorials Permutations Combinations O1 choose Ch1 questions N1 = C(2, 2) O2 choose Ch2 questions N2 = C(3, 2) O3 choose Ch3 questions N3 = C(5, 2) O4 choose Ch4 questions N4 = C(4, 2) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 66. Factorials Permutations Combinations O1 choose Ch1 questions N1 = C(2, 2) O2 choose Ch2 questions N2 = C(3, 2) O3 choose Ch3 questions N3 = C(5, 2) O4 choose Ch4 questions N4 = C(4, 2) So by the multiplication principle: C(2, 2)C(3, 2)C(5, 2)C(4, 2) = 1(3)(10)(6) = 180 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 67. Factorials Permutations Combinations (b) Suppose she writes the final by (1) choosing a set of 8 questions as above, and then (2) randomly ordering those 8 questions. Accounting for both operations, how many distinct final exams are possible? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 68. Factorials Permutations Combinations (b) Suppose she writes the final by (1) choosing a set of 8 questions as above, and then (2) randomly ordering those 8 questions. Accounting for both operations, how many distinct final exams are possible? O1 choose questions as in (a) N1 = 180 O2 put these questions in random order university-logo Jason Aubrey Math 1300 Finite Mathematics
- 69. Factorials Permutations Combinations (b) Suppose she writes the final by (1) choosing a set of 8 questions as above, and then (2) randomly ordering those 8 questions. Accounting for both operations, how many distinct final exams are possible? O1 choose questions as in (a) N1 = 180 O2 put these questions in random order N2 = P(8, 8) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 70. Factorials Permutations Combinations (b) Suppose she writes the final by (1) choosing a set of 8 questions as above, and then (2) randomly ordering those 8 questions. Accounting for both operations, how many distinct final exams are possible? O1 choose questions as in (a) N1 = 180 O2 put these questions in random order N2 = P(8, 8) So, by the multiplication principle: 180 · P(8, 8) = 180 · 8! = 7, 257, 600 university-logo Jason Aubrey Math 1300 Finite Mathematics
- 71. Factorials Permutations Combinations (c) Suppose the questions must be in order by chapter. For example, the Chapter 1 questions must come before the Chapter 2 questions, the Chapter 2 questions must come before the Chapter 3 questions, etc. In this case, how many different final exams are possible? university-logo Jason Aubrey Math 1300 Finite Mathematics
- 72. Factorials Permutations Combinations (c) Suppose the questions must be in order by chapter. For example, the Chapter 1 questions must come before the Chapter 2 questions, the Chapter 2 questions must come before the Chapter 3 questions, etc. In this case, how many different final exams are possible? O1 choose Ch1 questions N1 = P(2, 2) O2 choose Ch2 questions N2 = P(3, 2) O3 choose Ch3 questions N3 = P(5, 2) O4 choose Ch4 questions N4 = P(4, 2) university-logo Jason Aubrey Math 1300 Finite Mathematics
- 73. Factorials Permutations Combinations (c) Suppose the questions must be in order by chapter. For example, the Chapter 1 questions must come before the Chapter 2 questions, the Chapter 2 questions must come before the Chapter 3 questions, etc. In this case, how many different final exams are possible? O1 choose Ch1 questions N1 = P(2, 2) O2 choose Ch2 questions N2 = P(3, 2) O3 choose Ch3 questions N3 = P(5, 2) O4 choose Ch4 questions N4 = P(4, 2) So, by the multiplication principle: P(2, 2)P(3, 2)P(5, 2)P(4, 2) = 2, 880 university-logo Jason Aubrey Math 1300 Finite Mathematics