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Fall 2014 COMP 2300 Discrete Structures for Computation PowerPoint Presentation

Fall 2014 COMP 2300 Discrete Structures for Computation

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### Fall 2014COMP 2300 Discrete Structures for Computation

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

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

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

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

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

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

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

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

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

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

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

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

Possibility Trees and the Multiplication Rule

Donghyun (David) Kim

Department of Mathematics and PhysicsNorth Carolina Central University

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

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

- 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

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

- 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.

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

- 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

- 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

- 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 Ann became a president
- 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?

- 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)

- 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.

- 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!.

- 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

- 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?

- 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.

- 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

- 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

- How many different ways can three of the letters of the word BYTES be chosen and written in a row?
- How many different ways can this be done if the first letter must be B?

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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