Computing Fundamentals 2 Lecture 5 Combinatorial Analysis

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Computing Fundamentals 2Lecture 5 Combinatorial
Analysis

• Lecturer Patrick Browne
• http//www.comp.dit.ie/pbrowne/
• Room K408
• Based on Chapter 16.
• A Logical approach to Discrete Math
• By David Gries and Fred B. Schneider

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Combinatorial Analysis

• Counting
• Permutations
• Combinations
• The Pigeonhole Principle
• Examples

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Combinatorial Analysis

• Combinatorial analysis deals with permutations of
  a set or bag and combinations of a set, which
  lead to binomial coefficients and the Binomial
  Theorem.

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Rules of Counting

• Rule of sums The size of the union on n finite
  pair wise disjoint sets is the sum of their
  sizes.
• Rule of product The size of the cross product of
  n sets is the product of their sizes.
• Rule of difference The size of a set with a
  subset removed is the size of the set minus the
  size of the subset.

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Product Rule Example

• If each license plate contains 3 letters and 2
  digits. How many unique licenses could there be?
  
• Using the rule of products.
• 26 ? 26 ? 26 ? 10 ? 10 1,757,600

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Permutation of a set

• A permutation of a set of elements is a linear
  ordering (or sequence) of the elements e.g.
• 1,4,5
• Permutation 1 1, 4, 5
• Permutation 2 1, 5, 4
• An anagram is a permutation of words.
• There are n ? (n 1) ? (n - 2) .. 1 permutations
  of a set of n elements.
• This is factorial n, written n!

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Calculating Factorial

• module FACT
• protecting(INT)
• -- Two notations for factorial
• op _! Nat - NzNat prec 10
• op fact Nat - NzNat
• var N Nat
• -- Notation 1
• eq 0 ! 1 .
• ceq N ! N (N - 1) ! if N 0 .
• -- Notation 2
• eq fact(0) 1 .
• ceq fact(N) N fact(N - 1) if N 0 .
• open FACT
• red 4 ! .
• red fact(4) .

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Permutation of a set

• Sometimes we want a permutation of size r from a
  set of size n.
• (16.4) P(n,r) n!/(n-r)!
• The number of 2 permutations of BYTE is
• P(4,2) 4!/(4-2)! 4 ? 3 12
• BY,BT,BE,YB,YT,YE,TB,TY,TE,EB,EY,ET
• P(n,0) 1
• P(n,n-1) P(n,n) n!
• P(n,1) n

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Calculating Permutations and Combinations of sets

• mod CALC
• pr(FACT)
• op permCalc Int Int - Int
• op combCalc Int Int - Int
• vars N R Int
• -- Compute permutation where order matters abc
  / bac
• -- A permutation is an ordered combination.
• -- perm calculates how many ways R items can be
  selected from N items
• eq permCalc(N , R) fact(N) quo fact(N - R) .
• -- combination of N things taking R at a time
• -- Note extra term in divisor.
• eq combCalc(N , R) fact(N) quo (fact(N - R)
  fact(R)) .
• open CALC
• -- Permutation from 10 items taking 7 at a time
• red permCalc(10,7) . gives 604800
• -- Combination from 10 items taking 7 at a time

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Permutation with repetition of a set

• An r-permutations is a permutation that allows
  repetition. Here are all the 2-permutation of the
  letters in SON SS,SO,SN,OS,OO,ON,NS,NO,NN.
• Given a set of size n, in constructing an
  r-permutation with repetition, for each element
  we have n choices.
• (16.6) The number of r permutations with
  repetition of a set of size n is nr, repetition
  is allowed in the permutation not in the original
  set.

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Permutation of a bag

• A bag may have duplicate elements.
• Transposition of equal (or duplicate) elements in
  a permutation does not yield a different
  permutation e.g. AAAA.
• Hence, there will be fewer permutations of a bag
  than a set of the same size. The permutations on
  the set S,O,N and the bag ?M,O,M? are
• S,O,N SON,SNO,OSN,ONS,NSO,NOS
• ?M,O,M? MOM,MMO,OMM

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Permutation of a bag General Rule

• (16.7) The number of permutations of a bag of
  size n with k distinct elements occurring n1, n2,
  n3,.. nk times is
• n!
• n1! ? n2! ? n3! ... ? nk!

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Permutation of a bag

• Calc. size of S,O,N example
• red permCalc(3,3) gives 6
• Calc. size of MOM example
• red fact(3) quo (fact(1) fact(2)) .
• O occurs once, M twice, gives 3

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Permutation of a bag

• Consider the permutation of the 11 letters of
  MISSISSIPPI. M occurs 1 time, I occurs 4 times, S
  occurs 4 times, and P occurs 2 times.
• red fact(11) quo
• (fact(1) fact(2) fact(4) fact(4)) .

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Permutation of a bag

• ?O? a single permutation
• ?M1,O, M2? , label the two copies of M.
• We can distinguish the Ms.
• M1M2O,M2M1O,M1OM2,M2OM1,OM1M2,OM2M1,

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Example Combinations of a Set

• An r-combination of a set is a subset of size r.
  A permutation is a sequence while a combination
  is a set.
• The 2-permutations (seq.) of SOHN is
• SO,SH,SN,OH,ON,OS,HN,HS,HO,NS,NO,NH
• The 2-combinations (set) of SOHN is
• S,O,S,H,S,N,O,H,O,N,H,N

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Combinations of a Set

• The binomial coefficient, n choose r is
  written

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Pascals Triangle
Beginning with row 0 and place 0, the number 20
appears in row 6, place 3. In CafeOBJ we can
check this. red combCalc(6,3) . gives 20 red
combCalc(7,4) . gives 35 red combCalc(7,3) .
gives 35 See web page http//cob.isu.edu/parkerKR
/courses/CIS220/Programs/P9_pascalsTriangle.htm
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Special Combinations of a Set
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Calculating factorial and division
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Calculating "n choose k".
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Combinations of a Set

• (16.10) The number of r-combinations of n
  elements is
• A student has to answer 6 out of 9 questions on
  an exam. How many ways can this be done?

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Combinations with repetitions of a Set

• An r-combination with repetitions of a set S of
  size n is a bag of size r all of whose elements
  are in S. An r-combination of a set is a subset
  of that set an r-combination with repetition of
  a set is a bag, since its elements need not be
  distinct.

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Combinations with repetitions of a Set

• For example, the 2-combinations with repetition
  of SON are the bags
• ?S,O?,?S,N?,?O,N?,?S,S?,?O,O?,?N,N?
• On the other hand, the 2-permutations with
  repetition are the sequences
• SS,SO,SN,OS,OO,ON,NS,NO,NN

Note SO and OS are distinct permutations
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Combinations with repetitions of a Set

• (16.12) The number of r-combinations with
  repetition of a set of size n is

Combination size
Repetitions size
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Combinations with repetitions of a Set

• Suppose 7 people each gets either a burger, a
  cheese burger, or fish (3 choices). How many
  different orders are possible? The answer is the
  number of 7-combinations with repetition of a set
  of 3 elements.

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The Equivalence of three statements

• (16.13) The following numbers are equal
• The number of integer solutions of the equation
  x1x2x3...xnr.
• The number of r-combinations with repetition of a
  set of size n.
• The number of identical ways r identical objects
  can be distributed among n different containers.

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Rule of sum and product

• A class has 55 boys and 56 girls. What is the
  total number of students in the class, and how
  many different possible boy girl pairs are there?
• Two disjoint sets, boys and girls, rule of sum
  implies 5556111 students. The rule of product
  says 55 ? 56 3080.

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Rule of sum and product

• A student can pass the language requirement on a
  course by
• (i) gaining proficiency in French, German, or
  Japanese.
• (ii) gaining minimal qualification, which
  involves two semester of (ii)(a) German,
  Japanese, or Italian and (ii)(b) two semesters of
  Korean or Hindi.
• In how many different ways can the language
  requirement be satisfied?

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Rule of sum and product

• The set P of ways in which proficiency can be
  gained has 3 elements. Let the set S represent
  the way in which minimal qualification can be
  satisfied. Each element of S is a pair whose
  first element is either French, German, Japanese,
  or Italian and whose second element is either
  Korean or Hindi. The rule of products gives
  S4?28. Adding these using the rule of sums
  gives PS3811

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Rule of sum and product

• In how many different ways can the language
  requirement be satisfied?
• Proficiency 3
• MinimalQualification 4 ? 2 8
• P M 8 3 11

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Rule of sum and product

• One bag contains a red ball and a black ball (2).
  A second bag contains a red ball, a green ball,
  and a blue ball (3). A person randomly picks
  first a bag and then a ball. In what fraction of
  cases will a red ball be selected?
• PossibleSelections 23 5
• PossibleRed 2
• Fraction of red picked 2/5

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Permutations

• How many permutations of the letters are there in
  the following words
• LIE n3, 3! 6
• BRUIT n5, 5! 120
• CALUMMNY n7, 7!5040

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Permutations of a bag

• A coin is tossed 5 times, landing Head or Tails
  to form an outcome. One possible outcome is
  HHTTT.
• How many possible outcomes are there?
• How many outcomes have one Head?
• How many outcomes contain at most one Head?

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Permutations of a bag

• How many possible outcomes are there?
• Rule of product giving 2532 possible outcomes.

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Permutations of a bag

• How many outcomes have one Head?
• Permutation of a bag with 1 Head and four Tails.

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Permutations of a bag

• How many outcomes contain at most one Head?
• One Head
• No Heads
• At least one Head 1 5 6 (rule of sums)

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Combinations of Set

• A chairman has to select a committee of 5 from a
  facility of 25. How many possibilities are there?
  
• How many possibilities are there if the chair
  should be on the committee?

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The Pigeonhole Principle

• (16.43) If more than n pigeons are placed in n
  holes, at least one hole will contain more than
  one pigeon.
• With more than n pigeons in n holes the average
  number of pigeons per hole is greater than one.
• The statement at least one hole will contain
  more than one pigeon is equivalent to the
  maximum number of pigeons in any whole is greater
  than one.

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The Pigeonhole Principle

• Abstracting from pigeons and holes.
• Let av.S denote the average number of elements in
  bag S.
• Let max.S denote the maximum number of elements
  in bag S.
• av.S 1 implies max.S 1
• (16.45) Pigeonhole Principle.
• av.S ? max.S

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The Pigeonhole Principle

• (16.46) Pigeonhole Principle.
• av.S ? max.S
• Where
• 3.1 4 ceiling of real number

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Example The Pigeonhole Principle

• (16.47) Prove that in a room of eight people, at
  lease two of them have birthdays on the same day
  of the week.
• Let bag S contain, for each day of the week the
  number of people in the room whose birthday is on
  that day. The number of people is 8 the number of
  days is 7.

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Example The Pigeonhole Principle

• max.S
• ?
• av.S
• 8/7
• 2

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Example 2 The Pigeonhole Principle

• Suppose S is a set of six integers, each between
  1 and 12 inclusive. Prove that there must be two
  distinct nonempty subsets of S that have the same
  sum.
• Proof The sum of all the elements of S is at
  most 789101112 57. So the sum of the
  elements of any nonempty subset of S is at least
  1 and at most 57 there are 57 possibilities. But
  there are 261 63 nonempty subsets of S. Hence
  there must be two with the same sum.