Permutations and Combinations

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Permutations and Combinations
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Learning Objectives

• What are permutations.
• What are combinations.
• How to calculate binomial coefficients.
• What is the binomial theorem.
• Counting examples.

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Permutations and Combinations

• Urn models
• We are given set of n objects in an urn (dont
  ask why its called an urn - probably due to
  some statistician years ago) .
• We are going to pick (select) r objects from the
  urn in sequence. After we choose an object
• we can replace it-(selection with replacement)
• or not -(selection without replacement).
• If we choose r objects, how many different
  possible sequences of r objects are there?
• Does the order of the objects matter or not?

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Permutations and Combinations

• PermutationsSelection without replacement of r
  objects from the urn with n objects.
• A permutation is an arrangement.
• Order matters .
• After selecting the objects, two different
  orderings or arrangements constitute different
  permutations.

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Permutations and Combinations

• Choose the first object n ways,
• Choose the second object (since selection is
  without replacement) (n - 1) ways,
• ...
• Choose the rth object (n - r 1) ways.
• By the rule of product, The number of
  permutations of n things taken r at a time
• P(n,r) n(n - 1)(n - 2) . . . (n - r 1)
• Note

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Permutations and Combinations

• ExampleLet A and B be finite sets and let A
  ? B .Count the number of injections from A
  to B.Note there are no injections if A gt B
  (why?)
• There are P( B , A ) injections
• We order the elements of A, a1, a2, . . . and
  assume the urn contains the set B.
• There are B ways to choose the image of a1,
  B - 1 ways to choose the image of a2, and so
  forth.
• Selection is without replacement. Otherwise we do
  not construct an injection.

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Permutations and Combinations

• Combinations
• Selection is without replacement but order does
  not matter .
• It is equivalent to selecting subsets of size r
  from a set of size n.
• Divide out the number of arrangements or
  permutations of r objects from the set of
  permutations of n objects taken r at a time
• The number of combinations of n things taken r at
  a time

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Permutations and Combinations

• Other names for C(n, r)
• n choose r
• The binomial coefficient
• Example
• How many subsets of size r can be constructed
  from a set of n objects?
• The answer is clearly C(n, r) since once we
  select the objects (without replacement) the
  order doesn't matter.

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Permutations and Combinations

• CorollaryProof
• If we count the number of subsets of a set of
  size n, we get the cardinality of the power set.

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Permutations and Combinations

• ExampleSuppose you flip a fair coin n times.
  How many different ways can you get
• no heads? C(n, 0)
• exactly one head? C(n, 1)
• exactly two heads? C(n, 2)
• exactly r heads? C(n, r)
• at least 2 heads? 2 n - C(n, 0) - C(n, 1)

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Permutations and Combinations

• Pascal's IdentityProof
• We construct subsets of size k from a set with n
  1 elements given the subsets of size k and k-1
  from a set with n elements.
• The total will include
• all of the subsets from the set of size n which
  do not contain the new element C(n, k),
• plus
• the subsets of size k - 1 with the new element
  added C(n, k-1).

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Permutations and Combinations

• It produces

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Permutations and Combinations

• A good way to evaluate C(n, r) for large n and r
  (to avoid overflow).
• ExampleHow many bit strings of length 4 have
  exactly 2 ones (or exactly 2 zeros)?

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Permutations and Combinations

• AnalysisWe solve the problem by determining the
  positions of the two ones in the bit string.
• place the first one - 4 possibilities
• place the second one - 3 possibilities
• Hence it appears that we have (4)(3) 12
  possibilities.
• We enumerate them to make sure
• 0011, 0101, 1001, 0110, 1010, 1100.
• There are actually only 6 possibilities. What is
  wrong?

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Permutations and Combinations

• The answer would be correct if we had two
  different objects to place in the string.
• For example, if we were going to place an a and
  a b in the string we would have00ab, 00ba,
  0a0b, 0b0a, a00b, b00a,and so forth for a total
  of 12.
• But......the objects (1 and 1) are the same so
  the order is not important !
• Divide through by the number of orderings 2!
  2. Therefore the answer is 12/2 6.

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Permutations and Combinations

• ExampleHow many bit strings of length 4 have at
  least 2 ones?AnalysisTotal the number of
  strings that have
• zero 1s 1
• one 1 4Total 24 - 5 11.
• If the universe is the bit strings of length 4,
  what is the complement of the above set?
• What is its cardinality?

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Counting Examples

• An office building contains 27 floors and 37
  offices on each floor. How many offices are in
  the building ?
• Answer 27 37 999 (product rule)

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Counting Examples

• How many bit strings are there of length 6 or
  less ?
• Answer sum rule
• length 1 2
• length 2 4
• length 3 8
• length 4 16
• length 5 32
• length 6 64
• total 2 4 8163264 126

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Counting Examples

• How many strings of four decimal digits
• do not contain the same digit twice
• 10987 5040
• end with an even digit
• 1010105 5000
• have exactly three digits that are 9s
• 94 36

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Counting Examples

• How many different functions are there from a set
  with 10 elements to a set with
• one element 1
• two elements 2 10
• three elements 3 10

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Counting Examples

• How many bit strings of length seven begin with
  three 0s or end with three 1s?
• Principle of inclusion-exclusion
• begin with three 0s 2 4 16
• end with three 1s 2 4 16
• intersection 2
• total 16 16 - 2 30

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Pigeon-Hole Principle Examples

• Show that among any group of five (not
  necessarily consecutive integers), there are two
  with the same remainder when divided by 4.
• Pigeon-hole principle with 4 boxes (there are 4
  different remainders of division by 4)
• 41 5
• in any 5 integers, there are at least two with
  the same remainder when divided by 4

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Pigeon-Hole Principle Examples

• Suppose that there are 9 students in a class.
• Show that the class must have at least five male
  students or at least five female students.
• Generalized pigeon-hole principle 2 boxes,
  ?9/2? 5
• Show that the class must have at least three male
  students or at least seven female students.
• Either nb males gt2.
• Or nb males lt 2, and then nb females gt 7.

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Permutation and Combination

• List all permutations of a, b, c
• Permutations of a, b, c are bijections from and
  onto this same set. There are 3! 6
  permutations.
• a, b, c a, c, b b, a, c b, c, a
  c, a, b c, b, a

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Permutation and Combination

• A group contains n men and n women. How many ways
  are there to arrange these people in a row if the
  men and women alternate ?M W M W for M
  n ! P(n, n) for W n ! P(n,
  n) sub-total n! 2WMWM same n! 2total 2
  n! 2

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Permutation and Combination

• In how many ways can a set of five letters be
  selected from the English alphabet.
• We suppose that letters cannot be repeated
• If the order matters (such as in a word) P(26,
  5) 26.25.24.23.22 7893600Example abcdef,
  bacdef, ...
• If the order does not matter (such as in a
  set) C(26,5) P(26,5) / 5! 65780 Example
  a, b, c, d, e, f

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Permutation and Combination

• If letters can be repeated
• If the order matters 26 5 11881276
• If the order does not matter more complex
  C(26 5 - 1, 5) C(30, 5) 142506
• Number of groups of p objects (not necessarily
  different) with repetition taken from a set of m
  objects without taking the order into account
  C(m p - 1, p)

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Permutation and Combination

• A department contains 10 men and 15 women. How
  many ways are there to form a committee with 6
  members if it must have more women than men.
• Committees are not ordered and no duplicates are
  allowed.
• 6 women C(15, 6) . C(10,0)
• 5 women C(15,5) . C(10,1)
• 4 women C(15,4) . C(10,2)
• total (sum rule) C(15,6) C(15,5).10
  C(15,4).45

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Permutation and Combination

• Give the row of Pascals triangle immediately
  following 1 7 21 35 35 21 7 11 11 2
  11 3 3 11 4 6 4 11 5 10 10 5
  11 6 15 20 15 6 11 7 21 35 35 21 7
  1 1 8 28 56 70 56 28 8 1

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Permutation and Combination

• Develop (x y) 8 (x y)8 x8 8 x7 y 28
  x6 y2 56 x5 y3 70 x4 y4 56 x3 y5 28 x2
  y6 8 x y7 y8