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Chapter 9.2 Possibility Trees and the Multiplication Rule. Fall 2014 COMP 2300 Discrete Structures for Computation. Donghyun (David) Kim Department of Mathematics and Physics North Carolina Central University.

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Fall 2014 comp 2300 discrete structures for computation

Chapter 9.2

Possibility Trees and the Multiplication Rule

Fall 2014COMP 2300 Discrete Structures for Computation

Donghyun (David) Kim

Department of Mathematics and PhysicsNorth Carolina Central University


Possibilities for tournament play

  • Teams A and B are to play each other repeatedly until one wins two games in a row or a total of three games.

Possibilities for Tournament Play

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Possibilities for tournament play1

  • Teams A and B are to play each other repeatedly until one wins two games in a row or a total of three games.

    • How many ways can the tournament be played? 10

    • Assuming that all the ways of playing the tournament are equally likely, what is the probability that five games are needed to determine the tournament winner? 4/10

Possibilities for Tournament Play

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


The multiplication rule

  • If an operation consists of k steps and the first step can be performed in ways,the second step can be performed in ways,(regardless of how the first step was performed)the kth step can be performed in ways,(regardless of how the preceding steps was performed)

  • Example

    • What is the number of the possible outcomes of rolling 3 dices in a row? 6 x 6 x 6.

The Multiplication Rule

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Number of personal identification numbers pins

  • A typical PIN (personal identification number) is a sequence of any four symbols chosen from the 26 letters in the alphabet and the ten digits, with repetition allowed.

    • How many different PINs are possible? : 36 x 36 x 36 x 36 = 1,679,616 PINs.

    • If all PINs are equally likely, what is the probability that a PIN chosen at random contains no repeated symbol?: 36 x 35 x 34 x 33 / 36 x 36 x 36 x 36 = 0.8417

Number of Personal Identification Numbers (PINs)

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Counting the number of iterations of a nested loop

  • Consider the following nested loop: of any four symbols chosen from the 26 letters in the alphabet and the ten digits, with repetition allowed. How many times will the inner loop be iterated when the algorithm is implemented and run? 12

Counting the Number of Iterations of a Nested Loop

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


When the multiplication rule is difficult or impossible to apply

  • Three officers – a president, a treasurer, and a secretary – are to be chosen from among four people: Ann, Bob, Cyd, and Dan.

    • Suppose Ann became a president

      • Ann (P), Bob (T), Cyd (S) / Ann (P), Bob (S), Cyd(T)

      • Ann (P), Cyd (T), Dan (S) / Ann (P), Cyd(S), Dan (T)

      • Ann (P), Bob (T), Dan (S) / Ann (P), Bob (S), Dan (T)

    • # of possible choices for President = 4

      • # of possible cases: 4 * 6 = 24

  • Suppose that, for various reasons, Ann cannot be president and either Cyd or Dan must be secretary. How many ways can the officers be chosen?

When the Multiplication Rule Is Difficult or Impossible to Apply

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


When the multiplication rule is difficult or impossible to apply cont

  • Three officers – a president, a treasurer, and a secretary – are to be chosen from among four people: Ann, Bob, Cyd, and Dan.

  • Suppose that, for various reasons, Ann cannot be president and either Cyd or Dan must be secretary. How many ways can the officers be chosen? 8

    • # of possible choices for President = 3

    • Cyd is the president

      • Cyd (P), Ann (T), Bob (S) / Cyd (P), Ann (S), Bob (T)

      • Cyd(P), Ann (T), Dan (S) / Cyd (P), Ann (S), Dan (T)

      • Cyd(P), Bob (T), Dan (S) / Cyd (P), Bob (S), Dan (T)

    • Dan is the president

      • Dan (P), Ann (T), Bob (S) / Dan (P), Ann (S), Bob (T)

      • Dan (P), Ann (T), Cyd(S) / Dan (P), Ann (S), Cyd(T)

      • Dan (P), Bob (T), Cyd(S) / Dan (P), Bob (S), Cyd(T)

    • Bob is the president

      • Bob (P), Ann (T), Cyd(S) / Bob (P), Ann (S), Cyd(T)

      • Bob (P), Cyd(T), Dan (S) / Bob (P), Cyd(S), Dan (T)

      • Bob (P), Ann (T), Dan (S) / Bob (P), Ann (S), Dan (T)

When the Multiplication Rule Is Difficult or Impossible to Apply – cont’

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


More subtle use of the multiplication rule

  • Three officers – a president, a treasurer, and a secretary – are to be chosen from among four people: Ann, Bob, Cyd, and Dan.

  • Suppose that, for various reasons, Ann cannot be president and either Cyd or Dan must be secretary. How many ways can the officers be chosen?

    • Possible Secretary = 2 (Cyd or Dan)

    • Possible President = 2 (Bob or Dan if Cyd is P, Bod or Cyd if Dan is P)

    • Possible Treasuer = 2 (Reminders)

    • Total possible cases: 2 * 2 * 2 = 8.

More Subtle Use of the Multiplication Rule

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Permutations

  • A – are to be chosen from among four people: Ann, Bob, permutation of a set of objects is an ordering of the objects in a row. For example, the set of elements a, b, and c has six permutations.

  • Theorem 9.2.2

    • For any integer n with , the number of permutations of a set with n elements is n!.

Permutations

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Permutations of the letters in a word

  • How many ways can the letters in the word – are to be chosen from among four people: Ann, Bob, COMPUTER be arranged in a row?

    • 8!

  • How many ways can the letters in the word COMPUTER be arranged if the letter CO must remain next to each other (in order) as a unit?

    • 7!

  • If letters of the word COMPUTER are randomly arranged in a row, what is the probability that the letters CO remain next to each other (in order) as a unit? 7!/8! = 1/8

Permutations of the Letters in a Word

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Permutations of objects around a circle

  • At a meeting of diplomats, the six participants are to be seated around a circular table.

  • Since the table has no ends to confer particular status, it doesn’t matter who sits in which chair. But it does matter how the diplomats are seated relative to each other.

  • In other words, two seatings are considered the same if one is a rotation of the other. How many different ways can the diplomats be seated?

Permutations of Objects Around a Circle

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Permutations of objects around a circle cont

  • Solution seated around a circular table.

    • Fix one diplomat and consider all possible arrangement for the rest of diplomats. Then, we have 5! cases.

    • If confused, consider a smaller instance.

Permutations of Objects Around a Circle – cont’

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Permutations of selected elements

  • Given the set { seated around a circular table. a, b, c}, there are six ways to select two letters from the set and write them in order.

  • Each such ordering of two elements of {a, b, c} is called a 2-permutationof {a, b, c}.

  • An r-permutation of a set of n elements is an ordered selection of r elements taken from the set of n elements. The number of r-permutations of a set of n elements is denoted

Permutations of Selected Elements

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Theorem 9 2 3

  • If seated around a circular table. n and r are integers and , then the number of r-permutations of a set of n elements is given by the formula

  • Example – Evaluating r-Permutations

    • Evaluate P(5, 2) = 20

    • How many 4-permutations are there of a set of seven objects? P(7, 4) = 840

    • How many 5-permutations are there of a set of five objects? P(5, 5) = 120

Theorem 9.2.3

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Permutations of selected letters of a word

Permutations of Selected Letters of a Word

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


Proving a property of p n r

Proving a Property of P(n,r)

Fall 2014 COMP 2300 Department of Mathematics and Physics Donghyun (David) Kim North Carolina Central University


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