JABATAN MATEMATIK, SAINS DAN KOMPUTER POLITEKNIK SEBERANG PERAI BASIC NUR AHDA BT AWALLUL AZMI NOORAZWANI BT ALIAS COUNTING RULES
BASIC COUNTING RULES Nur Ahda bt Awallul Azmi Noorazwani bt Alias 2023 [JABATAN MATEMATIK SAINS DAN KOMPUTER] ©All rights reserved for electronic, mechanical, recording, or otherwise, without prior permission in writing from Politeknik Seberang Perai.
All rights reserved No part of this publication may be translated or reproduced in any retrieval system, or transmitted in any form or by any means, electronic, mechanical, recording, or otherwise, without prior permission in writing from Politeknik Seberang Perai. Published by Politeknik Seberang Perai Jalan Permatang Pauh, 13500 Permatang Pauh Pulau Pinang Editor NUR AHDA BINTI AWALLUL AZMI NOORAZWANI BINTI ALIAS Content Reviewer NOR SAADAH BINTI MOHAMMAD SOWI Cover Designer AZILAH BINTI ABDUL RAHIM Tel : 04-538 3322 Fax : 04-538 9266 Email : [email protected] Website : www.psp.edu.my FB : politeknikseberangperai Ig : politeknikseberangperai e ISBN 978-967-2774-47-1 Nur Ahda Awallul Azmi, editor. Noorazwani Alias, editor. Basic Counting Rules / Editor Nur Ahda Binti Awallul Azmi, Noorazwani Binti Alias 2023 Politeknik Seberang Perai, Electronics books - - Basic counting principle - - Permutaion - - Combination - - Government publications - - Malaysia - -
Acknowledgment In the name of Allah, the Most Gracious, the Most Merciful. Alhamdulillah. All praises to Allah for the strengths and His blessing in completing this ebook, Basic Counting Rules. We would like to express our deepest gratitude to our department, Jabatan Matematik Sains dan Komputer, Politeknik Seberang Perai for the opportunity and encouragement given for us to be the writer for this edition of e-book. Also, we would like to express our appreciation to all friends and all persons who helped us in completing this e-book whose names cannot be mentioned one by one for the help and support. The last but not the least, a special thanks go to our beloved family for their patience and understanding. All comments, suggestions and feedback will be appreciated. Nur Ahda bt Awallul Azmi Noorazwani bt Alias
Preface This e-book was created specifically for students who are taking Discrete Mathematics course. Counting is a fundamental concept that applied on many aspects of our lives. From calculating the number of items in a shopping cart to solving complex problems in mathematics and statistics, the ability to count is a skill that we often take for granted. This e-book will cover on the fundamental principles of counting, permutation and combination. Students will also learn how to apply these concepts in real-world scenarios. Examples and exercises will help students reinforce their understanding. By the end, students will be equipped with the confidence and competence to tackle various counting problems with ease. Happy counting!
Table of Content CHAPTER PAGES 01 Basic Counting Principle Sum Rule Product Rule Combination of Sum and Product Rule 1 2 - 4 4 - 6 6 - 7 02 Permutation What is Permutation? Permutation with Repetition Permutation without Repetition 9 10 11 - 17 17 - 29 03 Combination What is Combination? Combination with Repetition Combination without Repetition 32 33 33 - 35 35 - 37
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 1 01 BASIC COUNTING PRINCIPLE Counting principles are fundamental concepts used in combinatorics, the study of counting in mathematics. These principles help us to determine the number of possible outcomes in a given situation for example counting the number of ways to arrange objects or make choices. The two primary counting principles are the addition rule and multiplication rule.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 2 1.1 Sum Rule The sum rule, also known as addition rule is based on the idea that when events cannot occur together, we simply add up the number of choices for each event to find the total number of outcomes. The sum rule states that if one event can occur in m ways and another event can occur in n ways, then there are m + n ways that one of these events can occur. This rule can be applied to various situations, such as selecting objects from different sets, making choices between alternatives, or considering different cases. Let’s start with a simple example; Apply when making m choice OR n choice Keyword: OR m + n choices
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 3 Example 1 A student can choose one final project from one of the three lists. The three lists contain 12, 32 and 10 projects. How many possible projects are there to choose from? Solution: The student must choose only one project among all projects from three lists. Therefore, by using sum rule = 12 + 32 + 10 = 54 Example 2 You need to travel from Penang to Johor. You can either take the flight, bus, or by driving your own car. If there are 6 flights and 3 buses available for you, find how many ways to go to Johor.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 2 Solution: This example shows that all events are impossible to occur simultaneously. A person can only ride one transport at once. Therefore, by using sum rule = 6 + 3 + 1 = 10 To apply the sum rule, we must ensure that the events we count cannot happen at the same time. If there is any simultaneous occurrence, we need to use different counting principles. 1.2 Product Rule The product rule, also known as multiplication rule is used to determine the total number of outcomes when faced with multiple independent events. The product rule states that if one event can occur in m ways and another event can occur in n ways, then there are m x n ways that both events can occur together. This rule can be applied to various situations, such as arranging objects, selecting items from sets, or making sequential choices. Let’s start with a simple example; Apply when making m choice AND n choice Keyword: AND m x n choices
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 3 Example 3 We want to label the seat in an auditorium. The labels should combine one letter and followed by a number from 1 to 30. How many seats do we have to label? Solution: In the given situation, the label must include the first event, the letter and the second event, the number. Both events must occur together. The label should look like this. Therefore, product rule is applied in this example. Figure 1.1 Example of a label in an auditorium The letter in alphabet consists of 26 letters while the number is restricted from 1 to 30. = 26 × 30 = 780 Let’s try an example that related to a situation of selecting items. Example 4 Maya decided to buy a skirt and a blouse. The shop has 3 types of skirts and 4 types of blouse. In how many ways are there for Maya to select one skirt and one blouse?
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 4 Solution: The keyword here is AND. Maya will buy one skirt AND one blouse. This means that both of the items must occur at one time. According to the product rule, there are = 3 × 4 = 12 ways for Maya to select one skirt and one blouse 1.3 Combination of Sum and Product Rule Some of counting problems required BOTH rules to get the solution. For example, counting the possible number of passwords or PIN numbers. Example 5 A password can be length 6 characters long. The password must consist of letters or numbers from 1 to 9. How many different passwords can be created if i. the letter and number are allowed to be repeated in the password? ii. the letter and number are not allowed to be repeated in the password? Solution: i. The number of letters available are 26 (A to Z) while the numbers are 9 (1 to 9). For each character in the password, you can either choose a letter or a number.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 5 Therefore, for the first character, there are 26 + 9 = 35 choices. Same goes to the second, third until 6th character. Since all characters must come out together, the calculation should be like this = (26 + 9) × (26 + 9) × (26 + 9) × (26 + 9) × (26 + 9) × (26 + 9) = 1 838 265 625 of possible password can be created ii. The letter and number are not allowed to be repeated in the password? For the first character, there are 26 + 9 = 35 choices. However, the password has restriction where the letter and number are not allowed to be repeated, therefore, the second character should have 35 -1 = 34 choices. The calculation should be like this = 35 × 34 × 33 × 32 × 31 × 30 = 1 168 675 200 of possible passwords can be created
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 6 Tutorial / Exercise a. I have 3 red hats and 2 green hats. If I choose one hat, how many ways are there for me to make a selection? Answer: 5 b. An office building contains 37 floors and there are 40 offices on each floor. How many offices are there in that building? Answer: 1480 c. You are invited to try a 3-course set lunch at a restaurant. There are two sets to choose from, either Penang Nyonya Set or Western Set. Then, you can choose 1 appetizer, 1 main course and 1 dessert. How many ways for you to make a selection? Meal Penang Nyonya Set Western Set Appetizers 3 types 3 types Main course 2 types 2 types Desserts 3 types 2 types Answer: 30 d. One drawer contains 12 shirts and another drawer contains 7 neckties. How many ways to combine a shirt and a necktie? Answer: 84 e. The department will award a free computer to either a CS student or a CS professor. How many different choices are there if there are 530 students and 15 professors? Answer: 545
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 7 02 PERMUTATION Counting problems can be phrased in terms of an ordered arrangement of the object of a set with or without repetition. This arrangement is called permutation.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 8 2.1 What is Permutation? Permutation is an ordered arrangement of objects chosen from a set of objects. It is a selection where order or sequence matters. A simple example to show order matters is the password in a padlock. The combination lock can not be unlocked until the right sequence of digits or alphabets is not entered. Permutation in counting principle can be divided into two: with repetition and without repetition. Example 1 The three letters P, Q and R can be arranged in the following ways ✓ PQR ✓ PRQ ✓ QPR ✓ QRP ✓ RPQ ✓ RQP Each of the arrangement is called a permutation of the letters P, Q and R.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 9 3.2 Permutation with Repetition The selection rules are: 1 The order of arrangement is important 2 Distinct objects with repetition 3 Example PIN number, password, ranking number, arrangement of letters. Permutation of n different types of objects and repetition is allowed in the selection of r of them. × × × … … ( ) = where, n = the number of objects to choose from r = the number of choices we want to choose Example 2 A 3-digit code is made from a number 1, 2, 3, 4. How many such codes can be made if repetition of the given number is allowed?
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 10 Explanation: Calculation: = 4 × 4 × 4 = 64 You also can solve using this formula; = = 4 3 = 64 By using calculator fx-570ES PLUS to insert the value, tap on button Example 3 Penang International Airport assigns 3-letter codes to represent airport locations. How many airport codes are possible if repetition is allowed in forming this code? 1, 2, 3, 4 1. We need to make 3-digit code. 2. The first digit, we can put either number 1 or 2 or 3 or 4. 3. The second digit, we also can put either number 1 or 2 or 3 or 4. 4. The third digit, we also can put either number 1 or 2 or 3 or 4. 1, 2, 3, 4 1, 2, 3, 4 No of possibilities of first digit No of possibilities of second digit No of possibilities of third digit Where, n = total number from 1-4 and r = 3-digit code to made 4 3 =
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 11 Explanation: Calculation: = 26 × 26 × 26 = 17576 You also can solve using this formula; = = 263 = 17576 By using calculator fx-570ES PLUS to insert the value, tap on button 1. We need to make 3-letter codes. 2. The first letters, we can put either letter A until Z which is total letter is 26. 3. The second letters, we also can put either letter A until Z which is total letter is 26. 4. The third letters, we also can put either letter A until Z which is total letter is 26. No of possibilities of first letters No of possibilities of second letters No of possibilities of third letters 26 3 = 1st letter code 2nd letter code 3rd letter code Where, n = total letters from A to Z is 26 and r = 3-letters word to made
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 12 Example 4 A test consists of 5 multiple-choice questions. Each question has 3 possible answers. How many ways to answer all 5 questions? Explanation: There are 5 multiple-choice questions. Each question has 3 possible answers (assume 3 ways). So, we can conclude that 5 questions can be answered in 3 ways. Calculation: = 3 × 3 × 3 × 3 × 3 = 243 You also can solve using this formula; = = 3 5 = 243 By using calculator fx-570ES PLUS to insert the value, tap on button 3 ways 3 ways 3 ways 3 ways 3 ways 1st question can be answered in 3 ways 3 5 = 2nd question can be answered in 3 ways 3rd question can be answered in 3 ways 4th question can be answered in 3 ways 5th question can be answered in 3 ways Where, n = 3 possible answer and r = 5 multiple choice question
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 13 Permutation of multi-sets: Permutation of n different objects when P1 objects among n objects are similar, P2 objects of the second kind are similar, P3 objects of the third kind are similar, …., Pk objects of the kth kind are similar. ! 1! × 2! × 3! × … × ! Example 5 How many arrangements of the letters in the following words are possible? a. UNIFORM b. TEETH c. DISCRETE d. NINETEEN e. CALCULATOR a. UNIFORM Explanation: U N I F O R M Calculation: = 7! 1! × 1! × 1! × 1! × 1! × 1! × 1! = 5040 Total number of elements Number of each repeating element There are 7 letters in that word and no repeated letters.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 14 b. TEETH Explanation: T E E T H Calculation: = 5! 2! × 2! × 1! = 30 c. DISCRETE Explanation: D I S C R E T E Calculation: = 8! 2! × 1! × 1! × 1! × 1! × 1! × 1! = 20160 d. NINETEEN Explanation: N I N E T E E N Calculation: = 8! 3! × 3! × 1! × 1! = 1120 There are 5 letters in that word and letter E & T have repeated two times. There are 8 letters in that word and letter E have repeated two times. There are 8 letters in that word and letter E &N have repeated three times.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 15 e. CALCULATOR Explanation: C A L C U L A T O R Calculation: = 10! 2! × 2! × 2! × 1! × 1! × 1! × 1! = 453600 3.2 Permutation without Repetition The selection rules are: 1 The order of arrangement is important 2 Distinct objects without repetition 3 Example PIN number, password, ranking number, arrangement of letters and people. There are 10 letters in that word and letter A, L & C have repeated two times.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 16 Permutation of n different types of objects and repetition is not allowed in the selection of r of the, where ≤ , = ! ( − )! = × ( − 1) × ( − 2) × … × ( − + 1) Where n = the number of objects to choose from r = the number of choices we want to choose Example 6 A group of 4 students went for a dinner. However, there are only 3 seats to be filled. How many arrangements can we seat for the students? Explanation: From the total of 4 students, only 3 of them will be seated since there are 3 chairs provided. So, for the first chair we can choose one from 4 students to sit. Let say 1 student has been seated. So, there are 3 choices of students can be seated in the 2nd chair.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 17 2 students have been seated. So, there are 2 choice of students can be seated in the 3rd chair. Calculation: = 4 × 3 × 2 = 24 You also can solve using this formula; 4,3 = 4! (4 − 3)! = 24 By using calculator fx-570ES PLUS to insert the value, tap on button Example 7 There are six runners in a race. The winner receives a gold medal, the second-place finisher receives a silver medal and the third-place finisher receives a bronze medal. How many different ways are there to award these medals, if all possible outcomes of the race can occur and there are no ties? Explanation: From the total of 6 runners, only 3 of them will be choose as winner. There are 24 different ways to arrange the seat for 3 students out of 4 students. 4 shift × = Ρ 3
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 18 So, the gold medal, there are 6 possible runners to win. Let say 1 runner got gold medal, there are 5 possible runners to win. 2 runners already got gold and silver medal. So, there are 4 possible runners to win. Calculation: = 6 × 5 × 4 = 120 You also can solve using this formula; 6,3 = 6! (6 − 3)! = 120 By using calculator fx-570ES PLUS to insert the value, tap on button Example 8 How many 4 letter words can be arranged from the word “AEON BIG”. There are 120 different ways for runners to get the award. 6 shift × = Ρ 3
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 19 Explanation: From the total of 7 letters in the word “AEON BIG”, only 4 letters will be selected. How many choices of letter that can be arranged first, without repetition? Let say that letter A has been arranged. So, for the second place, there are 6 letters choice can be arranged. Let say that letter A & N has been arranged. So, for the third place, there are 5 letters choice can be arranged. Let say that letter A, N & G has been arranged. So, for the fourth place there are 4 letters choice can be arranged. Calculation: = 7 × 6 × 5 × 4 = 840 You also can solve using this formula; 7,4 = 7! (7 − 4)! = 840 7 letters choices There are 840 different ways to arrange the word “AEON BIG” 6 letters choices 5 letters choices 4 letters choices
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 20 By using calculator fx-570ES PLUS to insert the value, tap on button Example 9 A padlock has a 4-digit key code. All 4 digits in the key are different. What is the code for the padlock? Explanation: There are 10 numbers to choose from 0-9 to make 4-digit numbers in a padlock code without repetition. Let’s say that digit 7 has been put first in the 4-digit numbers. So, how many choices of digit left to be put next without repetition? Digit 7 and 5 has been put in the 4-digit numbers. So, how many choices of digit left to be put next without repetition? 7 shift × = Ρ 4 9 digit choices left 8 digit choices left 10 digit choices
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 21 Digit 7,5 and 3 have been put in the 4-digit numbers. So, how many choices of digit left to be put next without repetition? Calculation: = 10 × 9 × 8 × 7 = 5040 You also can solve using this formula; 10,4 = 10! (10 − 4)! = 5040 By using calculator fx-570ES PLUS to insert the value, tap on button Permutation of n different objects is n!, where; ! = × ( − 1) × ( − 2) × … × 2 × 1 Example 10 How many different words can be formed if the word contains all letters A, B, C, D, E and F? There are 5040 different ways to make 4-digit numbers in a padlock. 10 shift × = Ρ 4 7 digit choices left
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 22 Explanation: All letters A, B, C, D, E and F need to be formed without repetition. Let say letter A we choose first. So, next will be either letters B, C, D, E and F. Let say letter B we choose next. So, next will be either letters C, D, E and F. Let say letter C we choose next. So, next will be either letters D, E and F. Let say letter D we choose next. So, next will be either letters E and F. Let say letter E we choose next. So, next is F. The last letter is F. Calculation: ! = 6! = 720 By using calculator fx-570ES PLUS to insert the value, tap on button There are 720 different ways to form word from all letters A, B, C, D , E and F. 6 shift −1 = !
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 23 Example 11 There are 3 girls and 4 boys to be seated in a row of 7 chairs. In how many different ways they can be seated if there is no restriction? Explanation: All 3 women and 4 men will be seated in the chairs provided. Let say first women will be seated. So, next will be either 2 women and 4 men will be seated. they can Let say next chair will be seated by men. So, next will be either 2 women and 3 men will be seated. Let say next chair will be seated by men. So, next will be either 2 women and 2 men will be seated. Let say next chair will be seated by women. So, next will be either 1 woman and 2 men will be seated. Let say next chair will be seated by women. So, next will be either 2 men will be seated.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 24 Let say next chair will be seated by men. So, next will be either 1 man will be seated. Last chair will be seated by men. Calculation: ! = 7! = 5040 By using calculator fx-570ES PLUS to insert the value, tap on button Example 12 In how many ways can 5 people be lined up? Explanation: The 5 people are distinct. Once a person is in line, that person will not be repeated elsewhere in the line. Next line up will be 4 possible people in line. Next line up will be 3 possible people in line. There are 5040 different ways to be seated in a row of 7 chairs. 7 shift −1 = !
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 25 Next line up will be 2 possible people in line. Next line up will be 1 possible people in line. Calculation: ! = 5! = 120 By using calculator fx-570ES PLUS to insert the value, tap on button Example 13 How many permutations of the letters ABCDEFGH and ABC must be arranged side by side? Explanation: Because the letters ABC must arrange side by side, it will be considered as one object. We can find the answer by finding the number of permutations of six objects, namely the ABC and the individual letters D, E, F, G and H to be arrange. The letters ABC can also be arranged among themselves. There are 120 different ways to line up 5 people in a line. 5 shift −1 = ! ABC D E F G H Number of arrangements = 6! ABC Number of arrangements = 3!
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 26 Calculation: = 6! × 3! = 4320 By using calculator fx-570ES PLUS to insert the value, tap on button Example 14 Determine the number of ways in which letters of the word ‘DISCRETE’ can be arranged if the two E’s must not be together? Explanation: First, find the number of permutations of the word “DISRETE’. D I S C R E T E = 8! 2! × 1! × 1! × 1! × 1! × 1! × 1! = 20160 Next, find the number of permutations of two E’s together. The letter two E’s can also be arranged among themselves. EE There are 4320 different ways to arrange the letters. 6 shift −1 × ! 3 shift −1 ! = Number of arrangements = 7! There are 8 letters in that word and letter E have repeated two times. D I S C R T EE Number of arrangements = 2!
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 27 = 7! × 2! = 10080 By using calculator fx-570ES PLUS to insert the value, tap on button Therefore, the number of permutation if two E’s not to be together is; = - = 20160 − 10080 = 10080 The number of permutations of the word “DISRETE’ The number of permutations of two E’s together 7 shift −1 × ! 2 shift −1 ! =
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 28 Tutorial / Exercise a. Determine the number of 4-letter words that can be formed from the letters v, w, x, y and z in which the letters are allowed to be repeated. Answer: 625 b. In how many ways the 5 letters to be arranged from the word COMPUTER? Answer: 6720 c. How many arrangements can be formed from the words below: i. MARKET Answer: 720 ii. ECONOMICS Answer: 90720 iii. PROGRAMMING Answer: 4989600 d. A password consists of two letters of the alphabet followed by three digits chosen from 0 to 7. How many different possible password are there? Answer: 346112 e. Three married couples have bought 6 seats in the same row for a concert. In how many different ways can they be seated with no restrictions? Answer: 720 f. An interior decorator is trying to arrange a shelf containing eight books. In how many ways can she arrange the eight books? Answer: 40320 g. How many 3-digit arrangements can be formed from the digits 0,1,2,3,4,5 and 6 without repetition? Answer: 210 h. In a class of 20 students, 7 are to be chosen to join a choir competition. How many ways can I choose the students? Answer: 390700800
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 29 i. There are 7 horses in a race. How many different orders can the horses finish? Answer: 5040 j. How many permutations of the letters DOCUMENT and all vowel must be arranged together? Answer: 4320
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 30 03 COMBINATION Counting problems can be phrased in terms of an unordered arrangement of the object of a set with or without repetition. This arrangement is called combination.
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 31 3.1 What is Combination? Combination is an arrangement of objects chosen from a set of objects. It is a selection where order does not matter. In combination, you can select the objects in any order you like. A simple example to show order does not matter is the mix ingredients in a salad. It does not matter in what order you add all the ingredients. Eventually you will still produce a bowl of salad. Combination in counting principle can be divided into two: with repetition and without repetition. 3.2 Combination with Repetition The selection rules are: 1 The order of selection does not matter 2 Each object can be selected more than once 3 Example All items in general
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 32 The formula is as follow where represents the number of objects to choose from represents the number of choices we want to choose Example 1 A cookie shop has 4 flavours of cookies. If we want to buy 6 cookies, how many ways can we pick them? It does not matter in what order we pick the cookies. We may pick the chocolate first. Or we may pick the raisin first. And bear in mind that we may also pick all 6 cookies with the same flavours. Therefore, this situation is solved by using combination with repetition where +−1, = ( + − 1)! ! ( − 1)!
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 33 4+6−1,6 = (4 + 6 − 1)! 6! (4 − 1)! 9,6 = 9! 6! 3! 9,6 = 362880 4320 = 84 3.2 Combination without Repetition The selection rules are: 1 The order of selection does not matter 2 Each object can be selected only once 3 Example Person or car (item that have specific criteria)
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 34 The formula is as follow where represents the number of objects to choose from represents the number of choices we want to choose Example 2 How many different committees of three students can be formed from a group of four students? Let’s understand the situation. The committee will be consisting of 3 students chosen from a group of 4 students. There is no specific order to choose the students. This means that we can start by choosing STUDENT 1 followed by STUDENT 2 and finally STUDENT 3. Or we can also start by choosing STUDENT 3 followed by STUDENT 2 and finally STUDENT 1. No matter how we choose the students, the committee will still be able to form. Now let’s assume that we choose STUDENT 1 as the first member of committee. , = ! ! ( − )!
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 35 It is impossible to assign STUDENT 1 again to be the second member of the committee. Therefore, this situation is solved by using combination without repetition where 4,3 = 4! 3! 1! 9,6 = 24 6 = 4
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 36 Tutorial / Exercise a. I have 20 students in a class. I am going to pick 5 students for a prize. They will all get the same prize. How many ways can I choose the students? Answer: 15504 b. In how many ways can you choose 2 chocolates from a bag containing 6 different chocolates? Answer: 15 c. Ahmad wants to buy 4 pairs of shoes. If there are 5 colours to choose from, how many combinations of colours can he buy? Answer: 70 d. A meal at a Chinese restaurant allows you to choose 2 out of 8 side dishes. How many different combinations available if each side dishes can only be ordered once? Answer: 28
e B o o k P S P | B a s i c C o u n t i n g R u l e s | 37 References Faizah binti Abdul Muin, M. b. (2021). Basic Counting Rules. Balik Pulau: JMSK Politeknik Balik Pulau. Sullivan, M. (2008). Precalculus. Chicago State University: Pearson Education International. https://byjus.com/maths/permutation https://testbook.com/maths/permutation-withrepetition#:~:text=The%20Permutation%20with%20Repetition%20is,(r%20times). https://documents.uow.edu.au/content/groups/public/@web/@eis/@maas/documents /mm/uow168693.pdf
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