Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Math 1300: Section 7- 4 Permutations and Combinations

4,529 views

Published on

Published in: Education, Technology
  • Be the first to comment

Math 1300: Section 7- 4 Permutations and Combinations

  1. 1. Factorials Permutations Combinations Math 1300 Finite MathematicsSection 7-4: Permutations and Combinations Jason Aubrey Department of Mathematics University of Missouri university-logo Jason Aubrey Math 1300 Finite Mathematics
  2. 2. Factorials Permutations CombinationsProblem 1: Consider the set {p, e, n}. How many two-letter"words" (including nonsense words) can be formed from themembers of this set, if two different letters have to be used? university-logo Jason Aubrey Math 1300 Finite Mathematics
  3. 3. Factorials Permutations CombinationsProblem 1: Consider the set {p, e, n}. How many two-letter"words" (including nonsense words) can be formed from themembers 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. 4. Factorials Permutations CombinationsProblem 1: Consider the set {p, e, n}. How many two-letter"words" (including nonsense words) can be formed from themembers 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. 5. Factorials Permutations CombinationsProblem 1: Consider the set {p, e, n}. How many two-letter"words" (including nonsense words) can be formed from themembers 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. 6. Factorials Permutations CombinationsBoth problems involved counting the numbers of arrangementsof the same set {p, e, n}, taken 2 elements at a time, withoutallowing repetition. university-logo Jason Aubrey Math 1300 Finite Mathematics
  7. 7. Factorials Permutations CombinationsBoth problems involved counting the numbers of arrangementsof the same set {p, e, n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". university-logo Jason Aubrey Math 1300 Finite Mathematics
  8. 8. Factorials Permutations CombinationsBoth problems involved counting the numbers of arrangementsof the same set {p, e, n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. university-logo Jason Aubrey Math 1300 Finite Mathematics
  9. 9. Factorials Permutations CombinationsBoth problems involved counting the numbers of arrangementsof the same set {p, e, n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
  10. 10. Factorials Permutations CombinationsBoth problems involved counting the numbers of arrangementsof the same set {p, e, n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time. Thesecond example was concerned with the number ofcombinations of 3 objects taken 2 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
  11. 11. Factorials Permutations CombinationsDefinition (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. 12. Factorials Permutations CombinationsExamples 3! = 3(2)(1) = 6 university-logo Jason Aubrey Math 1300 Finite Mathematics
  13. 13. Factorials Permutations CombinationsExamples 3! = 3(2)(1) = 6 4! = 4(3)(2)(1) = 12 university-logo Jason Aubrey Math 1300 Finite Mathematics
  14. 14. Factorials Permutations CombinationsExamples 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. 15. Factorials Permutations CombinationsExamples 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. 16. Factorials Permutations CombinationsExamples 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. 17. Factorials Permutations CombinationsDefinition (Permutation of a Set of Objects)A permutation of a set of distinct objects is an arrangement ofthe objects in a specific order without repetition. university-logo Jason Aubrey Math 1300 Finite Mathematics
  18. 18. Factorials Permutations CombinationsTheorem (Number of Permutations of n Objects)The number of permutations of n distinct objects withoutrepetition, denoted by Pn,n is Pn,n = n(n − 1) · · · 2 · 1 = n! university-logo Jason Aubrey Math 1300 Finite Mathematics
  19. 19. Factorials Permutations CombinationsExample: Now suppose that the director of the art gallerydecides to use only 2 of the 4 available paintings, and they willbe arranged on the wall from left to right. We are now talkingabout 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. 20. Factorials Permutations CombinationsExample: Now suppose that the director of the art gallerydecides to use only 2 of the 4 available paintings, and they willbe arranged on the wall from left to right. We are now talkingabout 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. 21. Factorials Permutations CombinationsExample: Now suppose that the director of the art gallerydecides to use only 2 of the 4 available paintings, and they willbe arranged on the wall from left to right. We are now talkingabout 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. 22. Factorials Permutations CombinationsExample: Now suppose that the director of the art gallerydecides to use only 2 of the 4 available paintings, and they willbe arranged on the wall from left to right. We are now talkingabout 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. 23. Factorials Permutations CombinationsExample: Now suppose that the director of the art gallerydecides to use only 2 of the 4 available paintings, and they willbe arranged on the wall from left to right. We are now talkingabout 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. 24. Factorials Permutations CombinationsDefinition (Permutation of n Objects Taken r at a Time)A permutation of a set of n distinct objects taken r at a timewithout repetition is an arrangement of r of the n objects in aspecific order. university-logo Jason Aubrey Math 1300 Finite Mathematics
  25. 25. Factorials Permutations CombinationsTheorem (Number of Permutations of n Objects Taken r at aTime)The number of permutations of n distinct objects taken r at atime 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 nat 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. 26. Factorials Permutations CombinationsExample: Find P(5, 3) university-logo Jason Aubrey Math 1300 Finite Mathematics
  27. 27. Factorials Permutations CombinationsExample: 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. 28. Factorials Permutations CombinationsExample: 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 atime. university-logo Jason Aubrey Math 1300 Finite Mathematics
  29. 29. Factorials Permutations CombinationsExample: 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 atime.Example: Find P(5, 5), the number of permutations of 5objects taken 5 at a time. university-logo Jason Aubrey Math 1300 Finite Mathematics
  30. 30. Factorials Permutations CombinationsExample: 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 atime.Example: Find P(5, 5), the number of permutations of 5objects taken 5 at a time. P(5, 5) = 5(4)(3)(2)(1) = 120. university-logo Jason Aubrey Math 1300 Finite Mathematics
  31. 31. Factorials Permutations CombinationsExample: A park bench can seat 3 people. How many seatingarrangements are possible if 3 people out of a group of 5 sitdown? university-logo Jason Aubrey Math 1300 Finite Mathematics
  32. 32. Factorials Permutations CombinationsExample: A park bench can seat 3 people. How many seatingarrangements are possible if 3 people out of a group of 5 sitdown? P(5, 3) = (5)(4)(3) = 60. university-logo Jason Aubrey Math 1300 Finite Mathematics
  33. 33. Factorials Permutations CombinationsExample: A bookshelf has space for exactly 5 books. Howmany different ways can 5 books be arranged on thisbookshelf? university-logo Jason Aubrey Math 1300 Finite Mathematics
  34. 34. Factorials Permutations CombinationsExample: A bookshelf has space for exactly 5 books. Howmany different ways can 5 books be arranged on thisbookshelf? P(5, 5) = 5(4)(3)(2)(1) = 120 university-logo Jason Aubrey Math 1300 Finite Mathematics
  35. 35. Factorials Permutations CombinationsDefinition (Combination of n Objects Taken r at a Time)A combination of a set of n distinct objects taken r at a timewithout 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. 36. Factorials Permutations CombinationsTheorem (Number of Combinations of n Objects Taken r at aTime)The number of combinations of n distinct objects taken r at atime without repetition is given by: n Cn,r = r Pn,r = r! n! = 0≤r ≤n r !(n − r )! nNote: In place of the symbols Cn,r and , the symbols r university-logoC(n, r ) is often used. Jason Aubrey Math 1300 Finite Mathematics
  37. 37. Factorials Permutations CombinationsExamples: 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. 38. Factorials Permutations CombinationsExamples: 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. 39. Factorials Permutations CombinationsExample: In how many ways can you choose 5 out of 10friends to invite to a dinner party? university-logo Jason Aubrey Math 1300 Finite Mathematics
  40. 40. Factorials Permutations CombinationsExample: In how many ways can you choose 5 out of 10friends to invite to a dinner party?Does the order of selection matter? university-logo Jason Aubrey Math 1300 Finite Mathematics
  41. 41. Factorials Permutations CombinationsExample: In how many ways can you choose 5 out of 10friends 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. 42. Factorials Permutations CombinationsExample: In how many ways can you choose 5 out of 10friends 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. 43. Factorials Permutations CombinationsExample: How many 5-card poker hands will have 3 aces and2 kings? university-logo Jason Aubrey Math 1300 Finite Mathematics
  44. 44. Factorials Permutations CombinationsExample: How many 5-card poker hands will have 3 aces and2 kings?The solution involves both the multiplication principle andcombinations. university-logo Jason Aubrey Math 1300 Finite Mathematics
  45. 45. Factorials Permutations CombinationsExample: How many 5-card poker hands will have 3 aces and2 kings?The solution involves both the multiplication principle andcombinations. 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. 46. Factorials Permutations CombinationsExample: How many 5-card poker hands will have 3 aces and2 kings?The solution involves both the multiplication principle andcombinations. O1 : Choose 3 aces out of 4 N1 : C4,3 O2 : Choose 2 kings out of 4 N2 : C4,2Using 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. 47. Factorials Permutations CombinationsExample: A computer store receives a shipment of 24 laserprinters, including 5 that are defective. Three of these printersare selected to be displayed in the store. university-logo Jason Aubrey Math 1300 Finite Mathematics
  48. 48. Factorials Permutations CombinationsExample: A computer store receives a shipment of 24 laserprinters, including 5 that are defective. Three of these printersare selected to be displayed in the store.(a) How many selections can be made? university-logo Jason Aubrey Math 1300 Finite Mathematics
  49. 49. Factorials Permutations CombinationsExample: A computer store receives a shipment of 24 laserprinters, including 5 that are defective. Three of these printersare 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. 50. Factorials Permutations CombinationsExample: A computer store receives a shipment of 24 laserprinters, including 5 that are defective. Three of these printersare 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. 51. Factorials Permutations CombinationsExample: A computer store receives a shipment of 24 laserprinters, including 5 that are defective. Three of these printersare 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. 52. Factorials Permutations Combinations(b) How many of these selections will contain no defectiveprinters? university-logo Jason Aubrey Math 1300 Finite Mathematics
  53. 53. Factorials Permutations Combinations(b) How many of these selections will contain no defectiveprinters? P(19, 3) C(19, 3) = 3! university-logo Jason Aubrey Math 1300 Finite Mathematics
  54. 54. Factorials Permutations Combinations(b) How many of these selections will contain no defectiveprinters? P(19, 3) C(19, 3) = 3! 19 · 18 · 17 = 3·2·1 university-logo Jason Aubrey Math 1300 Finite Mathematics
  55. 55. Factorials Permutations Combinations(b) How many of these selections will contain no defectiveprinters? P(19, 3) C(19, 3) = 3! 19 · 18 · 17 = 3·2·1 = 969 university-logo Jason Aubrey Math 1300 Finite Mathematics
  56. 56. Factorials Permutations CombinationsExample: Suppose a certain bank has 4 branches inColumbia, 12 branches in St. Louis, and 10 branches inKansas City. The bank must close 6 of these branches. university-logo Jason Aubrey Math 1300 Finite Mathematics
  57. 57. Factorials Permutations CombinationsExample: Suppose a certain bank has 4 branches inColumbia, 12 branches in St. Louis, and 10 branches inKansas 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. 58. Factorials Permutations CombinationsExample: Suppose a certain bank has 4 branches inColumbia, 12 branches in St. Louis, and 10 branches inKansas 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. 59. Factorials Permutations CombinationsExample: Suppose a certain bank has 4 branches inColumbia, 12 branches in St. Louis, and 10 branches inKansas 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, 3branches in St. Louis, and 1 branch in Kansas City. In howmany ways can this be done? university-logo Jason Aubrey Math 1300 Finite Mathematics
  60. 60. Factorials Permutations CombinationsExample: Suppose a certain bank has 4 branches inColumbia, 12 branches in St. Louis, and 10 branches inKansas 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, 3branches in St. Louis, and 1 branch in Kansas City. In howmany 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. 61. Factorials Permutations CombinationsO1 : 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. 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. 63. Factorials Permutations CombinationsExample (FS09 Final Exam) During the semester, a professorwrote various problems to possibly include on the final exam ina math course. The course covered 4 chapters of a textbook.She wrote 2 problems from Chapter 1, 3 problems fromChapter 2, 5 problems from Chapter 3, and 4 problems fromChapter 4. She decides to put two questions from each chapteron the final exam. university-logo Jason Aubrey Math 1300 Finite Mathematics
  64. 64. Factorials Permutations CombinationsExample (FS09 Final Exam) During the semester, a professorwrote various problems to possibly include on the final exam ina math course. The course covered 4 chapters of a textbook.She wrote 2 problems from Chapter 1, 3 problems fromChapter 2, 5 problems from Chapter 3, and 4 problems fromChapter 4. She decides to put two questions from each chapteron the final exam.(a) In how many different ways can she choose two questionsfrom each chapter? university-logo Jason Aubrey Math 1300 Finite Mathematics
  65. 65. Factorials Permutations CombinationsO1 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. 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. 67. Factorials Permutations Combinations(b) Suppose she writes the final by (1) choosing a set of 8questions as above, and then (2) randomly ordering those 8questions. Accounting for both operations, how many distinctfinal exams are possible? university-logo Jason Aubrey Math 1300 Finite Mathematics
  68. 68. Factorials Permutations Combinations(b) Suppose she writes the final by (1) choosing a set of 8questions as above, and then (2) randomly ordering those 8questions. Accounting for both operations, how many distinctfinal 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. 69. Factorials Permutations Combinations(b) Suppose she writes the final by (1) choosing a set of 8questions as above, and then (2) randomly ordering those 8questions. Accounting for both operations, how many distinctfinal 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. 70. Factorials Permutations Combinations(b) Suppose she writes the final by (1) choosing a set of 8questions as above, and then (2) randomly ordering those 8questions. Accounting for both operations, how many distinctfinal 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. 71. Factorials Permutations Combinations(c) Suppose the questions must be in order by chapter. Forexample, the Chapter 1 questions must come before theChapter 2 questions, the Chapter 2 questions must comebefore the Chapter 3 questions, etc. In this case, how manydifferent final exams are possible? university-logo Jason Aubrey Math 1300 Finite Mathematics
  72. 72. Factorials Permutations Combinations(c) Suppose the questions must be in order by chapter. Forexample, the Chapter 1 questions must come before theChapter 2 questions, the Chapter 2 questions must comebefore the Chapter 3 questions, etc. In this case, how manydifferent 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. 73. Factorials Permutations Combinations(c) Suppose the questions must be in order by chapter. Forexample, the Chapter 1 questions must come before theChapter 2 questions, the Chapter 2 questions must comebefore the Chapter 3 questions, etc. In this case, how manydifferent 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

×