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**College of Information Technology & Design**

Discrete Mathematics Basics of Counting University of Jazeera College of Information Technology & Design Khulood Ghazal

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The product rule If there are n1 ways to do task 1, and n2 ways to do task 2 Then there are n1n2 ways to do both tasks in sequence This applies when doing the “procedure” is made up of separate tasks We must make one choice AND a second choice

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**Product rule example Total is 18 * 325 = 5850**

There are 18 math majors and 325 CS majors How many ways are there to pick one math major and one CS major? Total is 18 * 325 = 5850

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The sum rule If there are n1 ways to do task 1, and n2 ways to do task 2 If these tasks can be done at the same time, then… Then there are n1+n2 ways to do one of the two tasks We must make one choice OR a second choice

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**Sum rule example Total is 18 + 325 = 343**

There are 18 math majors and 325 CS majors How many ways are there to pick one math major or one CS major? Total is = 343

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**More complex counting problems**

We combining the product rule and the sum rule Thus we can solve more interesting and complex problems

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Example(1): The chairs of an auditorium are to be labeled with a letter and a positive integer not exceeding 100. What is the largest number of chairs that can be labeled differently?

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**What is the largest number of chairs that can be labeled differently?**

We can think of this problem as involving a sequence of two tasks: Assign a letter between A and Z Assign a number between 1 and 100 The Product Rule says that there are 26 * 100 = 2600 ways to do this. So we can label 2600 chairs.

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Example(2): How many different license plates are available if each plate contains a sequence of three letters followed by three digits? 26 choices for each letter 10 choices for each digit Total of: 26 * 26 * 26 * 10 * 10 * 10

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**Example(3): How many different bit strings are there of length seven?**

You probably already know it is 27. Think of this as: 2 (2 (2 (2 (2 (2 * 2)))))

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Example(4): How many different bit strings are there of length 1? Only 2: 0 or 1 How many different bit strings are there of length 2? There are 4: 00, 01, 10, 11 How many different bit strings are there of length 3? There are 8: 000, 001, 010, 011, 100, 101, 110, 111

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**Example(5): Each user on a computer system has a password**

Each password is six to eight characters long Each character is an uppercase letter or a digit Each password must contain at least one digit How many possible passwords are there?

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**There are 26 letters and 10 decimal digits = 36 characters that we can use to form passwords.**

For P6 (6-character) passwords, the Product Rule says there are 366 potential passwords. But passwords that are all letters are prohibited. There are 266 of these. So there are 366 – 266 for P6 passwords. Similarly, for P7 and P8 passwords. P7 = 367 – 267 P8 = 368 – 268 Total passwords = P6 + P7 + P8

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Example(6): How many bit strings of length eight either start with a 1 or end with the two bits 00? 1st Task: Construct a string beginning with a 1. 2nd Task: Construct a string ending with 00. Both tasks: Construct a string that begins with a 1 and ends with 00.

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1st: There are 28 ways to construct a binary string of 8 bits, but it starts with a 1, so there are 27 ways to construct an 8-bit binary string starting with 1. 2nd: Construct a string ending with 00. The product rule says there are 2 ways to choose the first 6 bits and 1 way to chose the last 2 bits, so there are 26 ways to construct this string. Both: Construct a string that begins with 1 and ends with 00. The product rule says there is 1 way to choose the first bit, 2 ways to chose the middle 5 bits, and 1 way to chose the last 2 bits, so there are 25 ways to construct this string. Total = ( ) – 25 = 160

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Example(7): A multiple choice test contains 10 questions. There are four possible answers for each question. How many ways can a student answer the questions on the test if every question is answered? 4*4*4*4*4*4*4*4*4*4 = 410 How many ways can a student answer the questions on the test if the student can leave answers blank? 5*5*5*5*5*5*5*5*5*5 = 510

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Section 4.1: The Basics of Counting As we have seen, one way to count the number of objects in a finite set S is to produce a one-to-one correspondence.

Section 4.1: The Basics of Counting As we have seen, one way to count the number of objects in a finite set S is to produce a one-to-one correspondence.

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