Generalized Permutations and Combinations Generating Permutations and Combinations

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Generalized Permutations and CombinationsGenerat
ing Permutations and Combinations
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Learning Objectives

• How to count permutations with repetition and
  combinations with repetition ?
• How to generate permutations and combinations ?

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Perm/Comb with Repetition

• Permutations with repetitions
• The number of r permutations with repetitions
  from a set with n elements is nr (The product
  rule).
• Combinations with repetitions
• The number of r combinations with repetitions
  from a set with n elements is C(n r 1,r)

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Perm/Comb with Repetition

• Proof
• Let A a1, , an. Consider the following bar
  code
• . . . We have exactly n 1 bars.
  These bars generate n holes. Place in hole
  number k, m stars to indicate that in the
  selection represented by this bar code you are
  going to pick m copies of ak. So the total number
  of symbols is going to be n 1 r and the
  number of r combinations with repetitions will
  thus be n 1 r choose r C(n r 1,n - 1)
  C(n r - 1, r)

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Perm/Comb with Repetition

• Comparison of counting methods for groups of
  objects

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Perm. Indistinguishable Objects

• Permutations with indistinguishable objectsThe
  number of permutations of n objects, where there
  are n1 indistinguishable objects of type 1, n2
  indistinguishable objects of type 2, and nk
  indistinguishable objects of type k are

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Perm. Indistinguishable Objects

• Proof place n objects
• n1 same objects can be placed in C(n, n1) ways
  (corresponds to choosing the positions among n)
• n2 same objects can be placed in C(n- n1, n2)
  ways
• nk same objects can be placed in C(n- n1- -
  nk-1, nk) ways
• product rule C(n, n1) C(n- n1, n2) C(n- n1-
  - nk-1, nk) n! / n1! n2! nk!

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Perm. Indistinguishable Objects

• ExampleHow many strings can be made up with
  all letters of the word EVERGREEN9! / 4! 1! 2!
  1! 1!

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Distributing Objects in Boxes

• Distributing n distinguishable objects into k
  different boxesni objects are placed in box i,
  i1, , k

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Distributing Objects in Boxes

• Proof
• n1 objects can be placed in box 1 in C(n, n1)
  ways (corresponds to choosing n1 objects among
  n)
• n2 objects can be placed in box 2 in C(n- n1, n2)
  ways
• nk objects can be placed in box k in C(n- n1- -
  nk-1, nk) ways
• product rule C(n, n1) C(n- n1, n2) C(n- n1-
  - nk-1, nk) n! / n1! n2! nk!

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Distributing Objects in Boxes

• ExampleHow many ways are there to distribute
  hands of 5 cards to four players from a standard
  deck of 52 cards ?52! / 5! 5! 5! 5! 32!

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Generating Permutations

• Lexicographic order (dictionary order)a1a2an
  precedes b1b2bn if, for some k, with 1 ? k ? n,
  a1 b1, a2 b2, a k-1 b k-1, and ak lt bk.
• Next permutation find k such that ak lt a
  k1and a k1 gt a k2 gt... gt a n

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Generating Permutations

• Examplepermutations of the set 1,2,3,4,5
  23415 precedes the permutation 23451
• Example 362541 --gt 364125

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Generating Permutations

• Algorithm principle a1a2an
• find the largest j such that aj lt aj1 and ? k
  gt j ak gt ak1
• put in jth position the least element in aj1an
  that is larger than aj (exchange with aj).
• arrange the elements after j in increasing order.
• Example 362541
• first pair from the right such that aj lt aj1 is
  (2,5)
• exchange 2 and 4
• reorder the other elements in increasing order
• result 364125

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Generating Permutations

• Example 45231
• first pair from the right such that aj lt aj1 is
  (2,3)
• exchange 2 and 3
• reorder the other elements in increasing order
• result 45312

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Generating Combinations

• A combination is a subset of a set of elements
  a1, a2, , an
• A combination can be represented as a bit string,
  where each bit represents an element in the set,
  and has the value 1 if the element is in the
  subset, and 0 if not.
• Example set of elements a, b, c, d, e, f
  subset a, b, f is represented by the bit
  string 110001, b, d is represented by the bit
  string 010100.

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Generating Combinations

• Finding the next combination is done by
  generating the next bit string (add 1 to the
  current bit string).Find the first 0 from the
  right, change it to a 1 and change all 1s to its
  right to 0s.
• Example 1000100the next combination will be
  1000101
• Example 101111the next combination will be
  110000

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Examples

• A croissant shop has plain croissants, cherry
  croissants, chocolate croissants, almond
  croissants, apple croissants and broccoli
  croissants. How many ways are there to choose
• a) a dozen croissants
• we suppose that there are at least 12 of each
  kind of croissants. C(6 12 - 1, 12) C(17,
  12)17! / 12! 5!
• plaincherrychocolatealmondapplebroccoli
• place 12 croissants in these boxes (5 bars and
  12 stars).
• There are 12 6 - 1 objects to choose from, and
  12 to choose from these.

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Examples

• b) three dozen croissants
• we suppose that there are at least 36 of each
  kind of croissants. C(36 6 - 1, 36) C(41,
  36)41! / 36! 5!
• plaincherrychocolatealmondapplebroccoli
• place 36 croissants in these boxes (5 bars and
  36 stars).
• There are 36 6 - 1 objects to choose from, and
  36 to choose from these.

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Examples

• c) two dozen croissants with at least two of
  each kind2 . 6 croissants are already attributed
  to their kind. There remains 12 croissants to
  choose from the 6 categories.
• C(12 6 - 1, 12) C(17, 12) 17! / 12! 5!
• d) two dozen croissants with no more than 2
  broccoli croissants
• 0 broccoli C(24 5 - 1, 24)
• 1 broccoli C(23 5 - 1, 23)
• 2 broccoli C(22 5 - 1, 22)
• total C(28, 24) C(27, 23) C(26, 22)

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Examples

• e) two dozen croissants with at least five
  chocolate croissants and at least three almond
  croissantsC(16 6 - 1, 16) C(21, 16) 21! /
  16! 5!
• f) two dozen croissants with at least one plain,
  two cherry three chocolate, one almond, two apple
  and no more than three broccoli 9 are already
  attributed.
• 0 broccoli C(15 5 - 1, 15)
• 1 broccoli C(14 5 - 1, 14)
• 2 broccoli C(13 5 - 1, 13)
• 3 broccoli C(12 5 - 1, 12)
• total C(19, 15) C(18, 14) C(17, 13) C(16,
  12)