Lec17_Combinatorics

# Lec17_Combinatorics - Discrete Structures TDS1191 - Lecture...

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[Lecture 17][Combinatorics] Discrete Structures 1 Discrete Structures TDS1191 - Lecture 17 - Combinatorics

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[Lecture 17][Combinatorics] Discrete Structures 2 The Sum Rule If a first task can be done in n 1 ways and a second task can be done in n 2 ways, and if these task cannot be done at the same time , then there are n 1 + n 2 ways to do either task . OR Suppose an event E can occur in m ways and a second event F can occur in n ways, and suppose both events cannot occur simultaneously, then E or F can occur in m + n ways. Basic Counting Techniques
[Lecture 17][Combinatorics] Discrete Structures 3 The Product Rule Suppose that a procedure can be broken down into two tasks . If there are n 1 ways to do the first task and n 2 ways to do the second tasks after the first task has been done, then there are n 1 n 2 ways to do the procedure. OR Suppose an event E can occur in m ways, and independent of this event, there is a second event F that can occur in n ways, Then combinations of E or F can occur in mn ways. Basic Counting Techniques

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[Lecture 17][Combinatorics] Discrete Structures 4 Question: A student can choose a computer project from one of three lists. The three lists contain 23, 15, and 19 possible projects respectively. How many possible projects are there to choose from? Solution: The student can choose a project from the first list in 23 ways, from the second list in 15 ways, and from the third lists in 19 ways. Hence, there are 23 + 15 = 19 = 57 projects to choose from. Basic Counting Techniques – Example 1
[Lecture 17][Combinatorics] Discrete Structures 5 Question: How many 5-digit numbers are there? Solution: There are 9 choices for the first digit, and 10 choices for each of the other 4 digits. Thus there are altogether 9 × 10 × 10 × 10 × 10 = 90,000 5-digit numbers. Basic Counting Techniques – Example 2

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[Lecture 17][Combinatorics] Discrete Structures 6 Question: A swimming team has 4 athletes who swim backstroke, 3 who swim breaststroke, 3 who swim butterfly, and 5 who swim freestyle. A medley relay consists of one swimmer from each of the four strokes. (i) How many choices does the coach have for a medley relay? (ii) How many choices does the coach have for a medley relay if one of the 3 athletes who swim butterfly is also one of the 5 athletes who swim freestyle? Solution: (i) By the product rule, the number of choices the coach have for a medley relay is 4 × 3 × 3 × 5 = 180. Basic Counting Techniques – Example 3 Solution: (i) No of choices = 4 x 3 x 3 x 5 = 180 (ii) We break the problem into 3 disjoint cases and use the Sum Rule. Let V denotes the versatile swimmer. Case 1: V swims butterfly: 4 × 3 × 1 × 4 = 48 Case 2: V swims freestyle: 4 × 3 × 2 × 1 = 24 Case 3: V swims neither: 4 × 3 × 2 × 4 = 96 Hence, the number of choices is 48 + 24 + 96 = 168.
[Lecture 17][Combinatorics] Discrete Structures 7 When two tasks can be done at the same time , we cannot use the sum rule to count the number of ways to do one of the

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## This note was uploaded on 02/03/2012 for the course IT 1191 taught by Professor Yong during the Spring '11 term at Multimedia University, Cyberjaya.

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Lec17_Combinatorics - Discrete Structures TDS1191 - Lecture...

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