IT IS PPT SLIDES CONTAINING TOPICS OF DISCRETE MATH AND GRAPH THEORY, Slides of Discrete Structures and Graph Theory

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CLASS-

MODULE-

Principles of Inclusion and Exclusion

DISCRETE MATHEMATICS AND GRAPH THEORY 25/01/

DISCRETE MATHEMATICS AND GRAPH THEORY 25/01/

 CONTENT

 Basics of Counting:

 The Sum Rule, the Product Rule

 Combinations with Repetition

 Pigeonhole Principle

 The Principle of Inclusion Exclusion

 The Generalizations of the Principles

 Derangements– Nothing is in its Right Place

 Rook Polynomials

 Arrangements with Forbidden Positions.

PRINCIPLES OF COUNTING Example 1: Suppose there are 16 boys and 18 girls in a class and we wish to select one of these students (either a boy or a girl) as the class representative. The number of ways of selecting a boy is 16 and the number of ways of selecting a girl is 18. Therefore, the number of ways of selecting a student (boy or girl) is 16 + 18 = 34. Example 2: Suppose a Hostel library has 12 books on Mathematics, 10 books on Physics, 16 books on Computer Science and 11 books on Electronics. Suppose a student wishes to choose one of these books for study. The number of ways in which he can choose a book is 12 + 10 + 16 + 11 = 49.

PRINCIPLES OF COUNTING The Product Rule Suppose that two tasks and are to be performed one after the other. If can be performed in different ways, and for each of these ways can be performed in different ways, then both of the tasks can be performed in different ways. More generally, suppose that tasks , , ……. are to be performed in a sequence. If can be performed in different ways and for each of these ways can be performed in different ways, and for each of different ways of performing and in that order, can be performed in different ways, and so on, then the sequence of tasks , , ……. can be performed in ……. different ways.

Example 6. Suppose a restaurant sells 6 South Indian dishes, 4 North Indian dishes, 3 hot beverages and 2

cold beverages, for breakfast, a student wishes to buy 1 South Indian dish and 1 hot beverage, or 1 North

Indian dish and 1 cold beverage.

SOLUTION : Then he can have the first choice in 6 × 3 = 18 ways and he can have the second choice in 4 × 2 = 8

ways. The total number of ways he can buy his breakfast items is 18 + 8 = 26.

Example 7. There are 20 married couple in a party. Find the number of ways of choosing one woman and

one man from the party such that the two are not married to each other.

SOLUTION : from the party, a woman can be chosen in 20 ways. Among the 20 men in the party, one is her

husband. Out of the 19 other men one can be chosen in 19 ways. Therefore, the required number is 20 × 19 = 380.

• (^) Example 8. A license plate consists of two English letters followed by four digits. If repetitions are allowed, how many of the plates have only vowels (A, E, I, O, U) and even digits (0, 2 , 4 , 6, 8)? SOLUTION : Each of the first two positions in a plate can be filled in 5 ways (with vowels and each of the remaining four places can be filled in 5 ways (with digits 0, 2, 4, 6, 8). Therefore, the number of possible license plates of the given type is (5 × 5) × (5 × 5 × 5 × 5) = 5 6 = 15,625.
• (^) Example 9. There are four bus routes between the places A and B and three bus routes between the places B and C. Find the number of ways a person can make a round trip from A via B if he does not use route more than once. SOLUTION : The person can travel from A to B in four ways and from B to C in three ways, but only two ways from C to B and only in three ways from B to A if he does not use a route more than once. Therefore, the number of ways he can make the round trip under the given condition is 4 × 3 × 2 × 3 = 72.

Example 11. Find the number of 3-digit even numbers with no repeated digits SOLUTION : Here we consider numbers of the form xyz, where each of x, y, z represents a digit under given restrictions. Since xyz has to be even, z has to be 0, 2, 4, 6, or 8. If z is 0, then x has 9 choices and if z is 2, 4, 6 or 8 ( choices) then x has 8 choices. (Note that x cannot be zero). Therefore, z and x can be chosen in (1 × 9) + (4 × 8) = 41 ways. For each of these ways, y can be chosen in 8 ways. Hence, the desired number is 41 × 8 = 328.

Example 12. Find the total number of positive integers that can be formed from the digits 1, 2, 3, 4 if no digit is repeated in any one integer. SOLUTION : We first note that no integer of the required type can contain more than 4 digits. Let s 1 , s 2 , s 3 , s 4 denote the number of integers of the required type containing one, two, three, four digits respectively.

• Since there are four digits, there are four integers containing exactly one digit (i.e. s 1 = 4), there are 4 × 3 = 12 integers containing exactly two digits (i.e. s 2 = 12), there are 4 × 3 × 2 = 24 integers containing exactly three digits (i.e. s 3 = 24) and there are 4 × 3 × 2 × 1 = 24 integers containing exactly three digits (i.e. s 4 = 24). Therefore, the required number is
• s 1 + s 2 + s 3 + s 4 = 64

The following are other interpretations of this number: _1) represents the number of ways in which r identical objects can be distributed among n distinct containers.

2. represents the number of non-negative integer solutions of the equation._ Example 1. A bag contains coins of seven different denominations, with at least one dozen coins in each denomination. In how many ways can we select a dozen coins from the bag?  (^) The selection consists in choosing with repetitions, r = 12 coins of n = 7 distinct denominations. The number of ways of making this selection is

Example 2. In how many ways can we distribute 10 identical marbles among 6 distinct containers?

• (^) The selection consists in choosing with repetitions, r = 10 marbles for n=6 distinct containers. The required number is Example 3. A cake shop sells 20 kinds of cakes. If there are at least a dozen cakes of each kind, in how many ways a dozen cakes can be chosen?
• (^) The required number is Example 4. Find the number of non-negative integer solutions of the equation  (^) The required number is

Example 7. Find the number of positive integer solutions of the equation we have Let us set The are all non-negative integers.

• (^) When written in terms of y’s, the given equation reads
• (^) The number of non-negative integer solutions of this equation is the required number. This number is

Example 8. Find the number of integer solutions of Where.  (^) Let us set Then are all non-negative integers. When written in terms of y’s, the given equation reads,

• (^) or
• (^) The number of non-negative integer solutions of this equation is the required number, and the number is

EXTENDED PIGEONHOLE PRINCIPLE: - Statement: - If pigeon are assigned to pigeonholes then one of the pigeonholes must contains at least pigeons.

PROBLEMS

1. Show that if you pick any five numbers from the integers 1-8, then two of them must add up to 9. Solution: - Let us first write all the numbers from 1 to 8 as (1,2,3,4,5,6,7,8) Now let's take any 5 numbers from 1 to 8 such as (1,3,4,7,8) As it given that any two of the numbers out of the 5 numbers we have chosen should be equal to sum 9. Let's add every two numbers so that we can get one such pair of numbers whose sum would be 9. Case 1>. 1 + 3 = 4 Case 2>. 3 + 4 = 7 Case 3>. 4 + 7 = 11 Case 4>. 7 + 8 = 15 Case 5>. 8 + 1 = [ 9 ]