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  1. 1. Permutations and CombinationsQuantitative Aptitude & Business Statistics
  2. 2. The Fundamental Principle of Multiplication• If there are• n1 ways of doing one operation,• n2 ways of doing a second operation, n3 ways of doing a third operation , and so forth, Quantitative Aptitude & Business 2 Statistics:Permutations and Combinations
  3. 3. • then the sequence of k operations can be performed in n1 n2 n3….. nk ways.• N= n1 n2 n3….. nk Quantitative Aptitude & Business 3 Statistics:Permutations and Combinations
  4. 4. Example 1• A used car wholesaler has agents who classify cars by size (full, medium, and compact) and age (0 - 2 years, 2- 4 years, 4 - 6 years, and over 6 years).• Determine the number of possible automobile classifications. Quantitative Aptitude & Business 4 Statistics:Permutations and Combinations
  5. 5. Solution 0-2 2-4 Full(F) 4-6 >6 0-2 2-4 Medium 4-6 >6 (M) 0-2 2-4 Compact 4-6 (C) >6The tree diagram enumerates all possibleclassifications, the total number of whichis 3x4= 12. Quantitative Aptitude & Business 5 Statistics:Permutations and Combinations
  6. 6. Example 2• Mr. X has 2 pairs of trousers, 3 shirts and 2 ties.• He chooses a pair of trousers, a shirt and a tie to wear everyday.• Find the maximum number of days he does not need to repeat his clothing. Quantitative Aptitude & Business 6 Statistics:Permutations and Combinations
  7. 7. Solution• The maximum number of days he does not need to repeat his clothing is 2×3×2 = 12 Quantitative Aptitude & Business 7 Statistics:Permutations and Combinations
  8. 8. 1.2 Factorials• The product of the first n consecutive integers is denoted by n! and is read as “factorial n”.• That is n! = 1×2×3×4×…. ×(n-1) ×n• For example,• 4!=1x2x3x4=24,• 7!=1×2×3×4×5×6×7=5040.• Note 0! defined to be 1. Quantitative Aptitude & Business 8 Statistics:Permutations and Combinations
  9. 9. •The product of any number ofconsecutive integers can beexpressed as a quotient of twofactorials, for example,• 6×7×8×9 = 9!/5! = 9! / (9 – 4)!• 11×12×13×14×15= 15! / 10! =15! / (15 – 5)!In particular,• n×(n – 1)×(n – 2)×...×(n – r + 1) Quantitative Aptitude & Business 9 Statistics:Permutations and Combinations
  10. 10. 1.3 Permutations• (A) Permutations• A permutation is an arrangement of objects.• abc and bca are two different permutations. Quantitative Aptitude & Business 10 Statistics:Permutations and Combinations
  11. 11. • 1. Permutations with repetition – The number of permutations of r objects, taken from n unlike objects, – can be found by considering the number of ways of filling r blank spaces in order with the n given objects. – If repetition is allowed, each blank space can be filled by the objects in n different ways. Quantitative Aptitude & Business 11 Statistics:Permutations and Combinations
  12. 12. 1 2 3 4 r n n n n n• Therefore, the number of permutations of r objects, taken from n unlike objects,• each of which may be repeated any number of times = n × n × n ×.... × n(r factors) = nr Quantitative Aptitude & Business 12 Statistics:Permutations and Combinations
  13. 13. 2. Permutations without repetition• If repetition is not allowed, the number of ways of filling each blank space is one less than the preceding one. 1 2 3 4 r n n-1 n-2 n-3 n-r+1 Quantitative Aptitude & Business 13 Statistics:Permutations and Combinations
  14. 14. Therefore, the number of permutations of r objects, taken from n unlike objects, each of which can only be used once in each permutation=n(n— 1)(n—2) .... (n—r + 1)Various notations are used to represent the number of permutations of a set of n elements taken r at a time; Quantitative Aptitude & Business 14 Statistics:Permutations and Combinations
  15. 15. • some of them are n P , Pr , P (n, r ) r n n! ( n − r )! Since n( n − 1)(n − 2)....(n − r + 1)(n − r )...3 ⋅ 2 ⋅ 1 = ( n − r )...3 ⋅ 2 ⋅ 1 Prn , n Pr , P (n, r ) = n( n − 1)(n − 2)....(n − r + 1) =P r n n!We have P = n (n − r )! r Quantitative Aptitude & Business 15 Statistics:Permutations and Combinations
  16. 16. Example 3• How many 4-digit numbers can be made from the figures 1, 2, 3, 4, 5, 6, 7 when• (a) repetitions are allowed;• (b) repetition is not allowed? Quantitative Aptitude & Business 16 Statistics:Permutations and Combinations
  17. 17. • Solution• (a) Number of 4-digit numbers = 74 = 2401.• (b) Number of 4 digit numbers =7 ×6 ×5 ×4 = 840. Quantitative Aptitude & Business 17 Statistics:Permutations and Combinations
  18. 18. Example 4• In how many ways can 10 men be arranged• (a) in a row,• (b) in a circle?• Solution• (a) Number of ways is = 3628800 10 P 10 Quantitative Aptitude & Business 18 Statistics:Permutations and Combinations
  19. 19. • Suppose we arrange the 4 letters A, B, C and D in a circular A arrangement as shown. D B• Note that the arrangements ABCD, BCDA, CDAB and C DABC are not distinguishable. Quantitative Aptitude & Business 19 Statistics:Permutations and Combinations
  20. 20. • For each circular arrangement there are 4 distinguishable arrangements on a line.• If there are P circular arrangements, these yield 4P arrangements on a line, which we know is 4!. 4! Hence P = = (4 − 1)!= 3! 4 Quantitative Aptitude & Business 20 Statistics:Permutations and Combinations
  21. 21. Solution (b)• The number of distinct circular arrangements of n objects is (n —1)!• Hence 10 men can be arranged in a circle in 9! = 362 880 ways. Quantitative Aptitude & Business 21 Statistics:Permutations and Combinations
  22. 22. (B) Conditional Permutations• When arranging elements in order , certain restrictions may apply.• In such cases the restriction should be dealt with first.. Quantitative Aptitude & Business 22 Statistics:Permutations and Combinations
  23. 23. Example 5 How many even numerals between 200and 400 can be formed by using 1, 2, 3, 4, 5 as digits (a) if any digit may be repeated; (b) if no digit may be repeated? Quantitative Aptitude & Business 23 Statistics:Permutations and Combinations
  24. 24. • Solution (a)• Number of ways of choosing the hundreds’ digit = 2.• Number of ways of choosing the tens’ digit = 5.• Number of ways of choosing the unit digit = 2.• Number of even numerals between 200 and 400 is 2 × 5 × 2 = 20. Quantitative Aptitude & Business 24 Statistics:Permutations and Combinations
  25. 25. •Solution (b)•If the hundreds’ digit is 2,then the number of ways of choosingan even unit digit = 1,and the number of ways of choosing atens’ digit = 3.•the number of numerals formed 1×1×3 = 3. Quantitative Aptitude & Business 25 Statistics:Permutations and Combinations
  26. 26. If the hundreds’ digit is 3, then the number of ways of choosing an even. unit digit = 2, and the number of ways of choosing a tens’ digit = 3.• number of numerals formed = 1×2×3 = 6.• the number of even numerals between 200 and 400 = 3 + 6 = 9 Quantitative Aptitude & Business 26 Statistics:Permutations and Combinations
  27. 27. Example 6In how many ways can 7 different books be arranged on a shelf (a) if two particular books are together; Quantitative Aptitude & Business 27 Statistics:Permutations and Combinations
  28. 28. • Solution (a)• If two particular books are together, they can be considered as one book for arranging.• The number of arrangement of 6 books = 6! = 720.• The two particular books can be arranged in 2 ways among themselves.• The number of arrangement of 7 books with two particular books Quantitative Aptitude & Business 28 Statistics:Permutations and Combinations
  29. 29. (b) if two particular books are separated?• Solution (b)• Total number of arrangement of 7 books = 7! = 5040.• the number of arrangement of 7 books with 2 particular books separated = 5040 -1440 = 3600. Quantitative Aptitude & Business 29 Statistics:Permutations and Combinations
  30. 30. (C) Permutation with Indistinguishable Elements• In some sets of elements there may be certain members that are indistinguishable from each other.• The example below illustrates how to find the number of permutations in this kind of situation. Quantitative Aptitude & Business 30 Statistics:Permutations and Combinations
  31. 31. Example 7In how many ways can the letters of the word “ISOS CELES” be arranged to form a new “word” ?• Solution• If each of the 9 letters of “ISOSCELES” were different, there would be P= 9! different possible words. Quantitative Aptitude & Business 31 Statistics:Permutations and Combinations
  32. 32. • However, the 3 S’s are indistinguishable from each other and can be permuted in 3! different ways.• As a result, each of the 9! arrangements of the letters of “ISOSCELES” that would otherwise spell a new word will be repeated 3! times. Quantitative Aptitude & Business 32 Statistics:Permutations and Combinations
  33. 33. • To avoid counting repetitions resulting from the 3 S’s, we must divide 9! by 3!.• Similarly, we must divide by 2! to avoid counting repetitions resulting from the 2 indistinguishable E’s.• Hence the total number of words that can be formed is 9! ÷3! ÷2! = 30240 Quantitative Aptitude & Business 33 Statistics:Permutations and Combinations
  34. 34. • If a set of n elements has k1 indistinguishable elements of one kind, k2 of another kind, and so on for r kinds of elements, then the number of permutations of the set of n elements is n! k1!k 2 !⋅ ⋅ ⋅ ⋅ k r ! Quantitative Aptitude & Business 34 Statistics:Permutations and Combinations
  35. 35. 1.4 Combinations• When a selection of objects is made with no regard being paid to order, it is referred to as a combination.• Thus, ABC, ACB, BAG, BCA, CAB, CBA are different permutation, but they are the same combination of letters. Quantitative Aptitude & Business 35 Statistics:Permutations and Combinations
  36. 36. • Suppose we wish to appoint a committee of 3 from a class of 30 students.• We know that P330 is the number of different ordered sets of 3 students each that may be selected from among 30 students.• However, the ordering of the students on the committee has no significance, Quantitative Aptitude & Business 36 Statistics:Permutations and Combinations
  37. 37. • so our problem is to determine the number of three-element unordered subsets that can be constructed from a set of 30 elements.• Any three-element set may be ordered in 3! different ways, so P330 is 3! times too large.• Hence, if we divide P330 by 3!,the result will be the number of unordered subsets of 30 elements taken 3 at a time. Quantitative Aptitude & Business 37 Statistics:Permutations and Combinations
  38. 38. • This number of unordered subsets is also called the number of combinations of 30 elements taken 3 at a time, denoted by C330 and 1 30 C = P3 30 3 3! 30! = = 4060 27!3!Quantitative Aptitude & Business Statistics:Permutations and Combinations 38
  39. 39. • In general, each unordered r- element subset of a given n- element set (r≤ n) is called a combination.• The number of combinations of n elements taken r at a time is denoted by Cnr or nCr or C(n, r) . Quantitative Aptitude & Business 39 Statistics:Permutations and Combinations
  40. 40. • A general equation relating combinations to permutations is 1 n n! C r = Pr = n r! (n − r )!r! Quantitative Aptitude & Business 40 Statistics:Permutations and Combinations
  41. 41. • Note:• (1) Cnn = Cn0 = 1• (2) Cn1 = n• (3) Cnn = Cnn-r Quantitative Aptitude & Business 41 Statistics:Permutations and Combinations
  42. 42. Example8• If 167 C 90+167 C x =168 C x then x is• Solution: nCr-1+nCr=n+1 Cr• Given 167 C90+167c x =168C x• We may write• 167C91-1 + 167 C91=167+1 C61• =168 C91• X=91 Quantitative Aptitude & Business 42 Statistics:Permutations and Combinations
  43. 43. Example9• If 20 C 3r= 20C 2r+5 ,find r• Using nCr=nC n-r in the right –side of the given equation ,we find ,• 20 C 3r =20 C 20-(2r+5)• 3r=15-2r• r=3 Quantitative Aptitude & Business 43 Statistics:Permutations and Combinations
  44. 44. Example 10• If 100 C 98 =999 C 97 +x C 901 find x.• Solution 100C 98 =999C 98 +999C97• = 999C901+999C97• X=999 Quantitative Aptitude & Business 44 Statistics:Permutations and Combinations
  45. 45. Example11• If 13 C 6 + 2 13 C5 +13 C 4 =15 C x ,the value of x is• Solution :• 15C x= 13C 6 + 13 C 5 + 13 C 4 =• =(13c 6+13 C 5 ) +• (13 C 5 + 13 C 4)• = 14 C 6 +14 C 5 =15C6• X=6 or x+6 =15• X=6 or 8 Quantitative Aptitude & Business 45 Statistics:Permutations and Combinations
  46. 46. Example12• If n C r-1=36 ,n Cr =84 and n C r+1 =126 then find r• Solution nCr 84 7 = = nCr −1 36 3• n-r+1 =7/3 * r• 3/2 (r+1)+1 =7/3 * r• nCr +1 126 3 r=3 = = nCr 84 2 Quantitative Aptitude & Business 46 Statistics:Permutations and Combinations
  47. 47. Example 13• How many different 5-card hands can be dealt from a deck of 52 playing cards? Quantitative Aptitude & Business 47 Statistics:Permutations and Combinations
  48. 48. Solution• Since we are not concerned with the order in which each card is dealt, our problem concerns the number of combinations of 52 elements taken 5 at a time.• The number of different hands is C525= 2118760. Quantitative Aptitude & Business 48 Statistics:Permutations and Combinations
  49. 49. Example 146 points are given and no three of them are collinear. (a) How many triangles can be formed by using 3 of the given points as vertices? Quantitative Aptitude & Business 49 Statistics:Permutations and Combinations
  50. 50. Solution:• Solution• (a) Number of triangles• = number of ways• of selecting 3 points out of 6• = C63 = 20. Quantitative Aptitude & Business 50 Statistics:Permutations and Combinations
  51. 51. • b) How many pairs of triangles can be formed by using the 6 points as vertices ? Quantitative Aptitude & Business 51 Statistics:Permutations and Combinations
  52. 52. • Let the points be A, B, C, D, E, F.• If A, B, C are selected to form a triangles, then D, E, F must form the other triangle.• Similarly, if D, E, F are selected to form a triangle, then A, B, C must form the other triangle. Quantitative Aptitude & Business 52 Statistics:Permutations and Combinations
  53. 53. • Therefore, the selections A, B, C and D, E, F give the same pair of triangles and the same applies to the other selections.• Thus the number of ways of forming a pair of triangles = C63 ÷ 2 = 10 Quantitative Aptitude & Business 53 Statistics:Permutations and Combinations
  54. 54. Example 15• From among 25 boys who play basketball, in how many different ways can a team of 5 players be selected if one of the players is to be designated as captain? Quantitative Aptitude & Business 54 Statistics:Permutations and Combinations
  55. 55. Solution• A captain may be chosen from any of the 25 players.• The remaining 4 players can be chosen in C254 different ways.• By the fundamental counting principle, the total number of different teams that can be formed is 25 × C244=265650. Quantitative Aptitude & Business 55 Statistics:Permutations and Combinations
  56. 56. (B) Conditional Combinations• If a selection is to be restricted in some way, this restriction must be dealt with first.• The following examples illustrate such conditional combination problems. Quantitative Aptitude & Business 56 Statistics:Permutations and Combinations
  57. 57. A committee of 3 men and 4 women is to beselected from 6 men and 9 women. If there is a married couple among the 15 persons, in how manyways can the committee be selected so that it contains the married Quantitative Aptitude & Business 57 Statistics:Permutations and Combinations
  58. 58. • Solution• If the committee contains the married couple, then only 2 men and 3 women are to be selected from the remaining 5 men and 8 women.• The number of ways of selecting 2 men out of 5 = C52 = 10. Quantitative Aptitude & Business 58 Statistics:Permutations and Combinations
  59. 59. • The number of ways of selecting 3 women out of 8 =C83 = 56.• the number of ways of selecting the committee = lO × 56 = 560. Quantitative Aptitude & Business 59 Statistics:Permutations and Combinations
  60. 60. Example 17• Find the number of ways a team of 4 can be chosen from 15 boys and 10 girls if (a) it must contain 2 boys and 2 girls, Quantitative Aptitude & Business 60 Statistics:Permutations and Combinations
  61. 61. • Solution (a)• Boys can be chosen in C152 = 105 ways• Girls can be chosen in C102 = 45 ways.• Total number of ways is 105 × 45 = 4725. Quantitative Aptitude & Business 61 Statistics:Permutations and Combinations
  62. 62. (b) it must contain at least 1 boy and 1 girl. • Solution : • If the team must contain at least 1 boy and 1 girl it can be formed in the following ways: • (I) 1 boy and 3 girls, with C151 × C103 = 1800 ways, • (ii) 2 boys and 2 girls, with 4725 ways, • (iii) 3 boys and 1 girl, with C153 × C101 = 4550 ways. Quantitative Aptitude & Business • the total number of teams is 62 Statistics:Permutations and Combinations
  63. 63. Example 18• Mr. .X has 12 friends and wishes to invite 6 of them to a party. Find the number of ways he may do this if (a) there is no restriction on choice, Quantitative Aptitude & Business 63 Statistics:Permutations and Combinations
  64. 64. • Solution (a) • An unrestricted choice of 6 out of 12 gives C126= 924. Quantitative Aptitude & Business 64 Statistics:Permutations and Combinations
  65. 65. two of the friends is a couple• (b) and will not attend separately, Quantitative Aptitude & Business 65 Statistics:Permutations and Combinations
  66. 66. B Solution• If the couple attend, the remaining 4 may then be chosen from the other 10 in C104 ways.• If the couple does not attend, then He simply chooses 6 from the other 10 in C106 ways.• total number of ways is C104 + C106 = 420. Quantitative Aptitude & Business 66 Statistics:Permutations and Combinations
  67. 67. Example 19Find the number of ways in which 30 students can be divided intothree groups, each of 10 students,if the order of the groups and the arrangement of the students in a group are immaterial. Quantitative Aptitude & Business 67 Statistics:Permutations and Combinations
  68. 68. • Solution• Let the groups be denoted by A, B and C. Since the arrangement of the students in a group is immaterial,• group A can be selected from the 30 students in C3010 ways . Quantitative Aptitude & Business 68 Statistics:Permutations and Combinations
  69. 69. • Group B can be selected from the remaining 20 students in C2010 ways.• There is only 1 way of forming group C from the remaining 10 students. Quantitative Aptitude & Business 69 Statistics:Permutations and Combinations
  70. 70. • Since the order of the groups is immaterial, we have to divide the product C3010 × C2010 × C1010 by 3!,• hence the total number of ways of forming the three groups is 1 × C3 × C10 × C10 30 20 10 3! Quantitative Aptitude & Business 70 Statistics:Permutations and Combinations
  71. 71. Example20• If n Pr = 604800 10 C r =120 ,find the value of r• We Know that nC r .r P r = nPr .• We will use this equality to find r• 10Pr =10Cr .r|• r |=604800/120=5040=7 |• r=7 Quantitative Aptitude & Business 71 Statistics:Permutations and Combinations
  72. 72. Example 21• Find the value of n and r• n Pr = n P r+1 and n C r = n C r-1Solution : Given n Pr = n P r+1 n –r=1 (i)n C r = n C r-1 n-r = r-1 (ii)Solving i and ii r=2 and n=3 Quantitative Aptitude & Business 72 Statistics:Permutations and Combinations
  73. 73. Multiple choice Questions Quantitative Aptitude & Business 73 Statistics:Permutations and Combinations
  74. 74. 1. Eleven students are participating in a race. In how many ways the first 5 prizes can be won?A) 44550B) 55440C) 120D) 90 Quantitative Aptitude & Business 74 Statistics:Permutations and Combinations
  75. 75. 1. Eleven students are participating in a race. In how many ways the first 5 prizes can be won?A) 44550B) 55440C) 120D) 90 Quantitative Aptitude & Business 75 Statistics:Permutations and Combinations
  76. 76. • 2. There are 10 trains plying between Calcutta and Delhi. The number of ways in which a person can go from Calcutta to Delhi and return• A) 99.• B) 90• C) 80• D) None of these Quantitative Aptitude & Business 76 Statistics:Permutations and Combinations
  77. 77. • 2. There are 10 trains plying between Calcutta and Delhi. The number of ways in which a person can go from Calcutta to Delhi and return• A) 99.• B) 90• C) 80• D) None of these Quantitative Aptitude & Business 77 Statistics:Permutations and Combinations
  78. 78. • 3. 4P4 is equal to• A) 1• B) 24• C) 0• D) None of these Quantitative Aptitude & Business 78 Statistics:Permutations and Combinations
  79. 79. • 3. 4P4 is equal to• A) 1• B) 24• C) 0• D) None of these Quantitative Aptitude & Business 79 Statistics:Permutations and Combinations
  80. 80. • 4.In how many ways can 8 persons be seated at a round table?• A) 5040• B) 4050• C) 450• D) 540 Quantitative Aptitude & Business 80 Statistics:Permutations and Combinations
  81. 81. • 4.In how many ways can 8 persons be seated at a round table?• A) 5040• B) 4050• C) 450• D) 540 Quantitative Aptitude & Business 81 Statistics:Permutations and Combinations
  82. 82. n n+1• 5. If P13 : P12 =3 : then 4 value of n is• A) 15• B) 14• C) 13• D) 12 Quantitative Aptitude & Business 82 Statistics:Permutations and Combinations
  83. 83. n n+1• 5. If P13 : P12 =3 : then 4 value of n is• A) 15• B) 14• C) 13• D) 12 Quantitative Aptitude & Business 83 Statistics:Permutations and Combinations
  84. 84. • 6.Find r if 5Pr = 60• A) 4• B) 3• C) 6• D) 7 Quantitative Aptitude & Business 84 Statistics:Permutations and Combinations
  85. 85. • 6.Find r if 5Pr = 60• A) 4• B) 3• C) 6• D) 7 Quantitative Aptitude & Business 85 Statistics:Permutations and Combinations
  86. 86. • 7. In how many different ways can seven persons stand in a line for a group photograph?• A) 5040• B) 720• C) 120• D) 27 Quantitative Aptitude & Business 86 Statistics:Permutations and Combinations
  87. 87. • 7. In how many different ways can seven persons stand in a line for a group photograph?• A) 5040• B) 720• C) 120• D) 27 Quantitative Aptitude & Business 87 Statistics:Permutations and Combinations
  88. 88. • 8. If 18 Cn = 18 Cn+ 2 then the value of n is ______A) 0B) –2C) 8D) None of above Quantitative Aptitude & Business 88 Statistics:Permutations and Combinations
  89. 89. • 8. If 18 Cn = 18 Cn+ 2 then the value of n is ______A) 0B) –2C) 8D) None of above Quantitative Aptitude & Business 89 Statistics:Permutations and Combinations
  90. 90. • 9. The ways of selecting 4 letters from the word EXAMINATION is• A) 136.• B) 130• C) 125• D) None of these Quantitative Aptitude & Business 90 Statistics:Permutations and Combinations
  91. 91. • 9. The ways of selecting 4 letters from the word EXAMINATION is• A) 136.• B) 130• C) 125• D) None of these Quantitative Aptitude & Business 91 Statistics:Permutations and Combinations
  92. 92. • 10 If 5Pr = 120, then the value of r is• A) 4,5• B) 2• C) 4• D) None of these Quantitative Aptitude & Business 92 Statistics:Permutations and Combinations
  93. 93. • 10 If 5Pr = 120, then the value of r is• A) 4,5• B) 2• C) 4• D) None of these Quantitative Aptitude & Business 93 Statistics:Permutations and Combinations
  94. 94. THE ENDPermutations and Combinations

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