SAJC 2021 H1 Math P & C notes

SAJC 2021 JC1 H1 Mathematics Chapter 1: Permutation and Combinations 0 Chapter 1 (Statistics) : Permutations and Combinations (Teacher’s Copy) Objectives: At the end of the chapter, you should be able to (a) understand and use the addition and multiplication principles for counting; (b) understand the concepts of permutation Pn r and combination Cn r , and differentiate between permutations and combinations; (c) understand the concept of arrangements of distinct objects in a line including cases involving restriction, and know how to calculate them Content 1.1 Basic Counting Principles 1.1.1 Addition Principle 1.1.2 Multiplication Principle 1.2 Permutations 1.3 Combinations

SAJC 2021 JC1 H1 Mathematics Chapter 1: Permutation and Combinations 1 1.1 BASIC COUNTING PRINC IPLES In the process of solving a counting problem, there are two very simple but basic principles that we always apply. They are called the Addition Principle and Multiplication Principle. 1.1.1 The Addition Principle (OR) Suppose you want to buy an ice cream. You have 5 flavours to choose from: Chocolate OR strawberry OR vanilla OR oreo OR mint. How many choices do you have? Answer: 1 + 1 + 1 + 1 + 1 = 5 Chocolate OR strawberry OR vanilla OR oreo OR mint In general, if there exists k non-overlapping categories of ways to perform an operation and the first category can be done in n1 ways, the second in n2 ways, …., the kth category in nk ways, then the operation can be done in 12 ... kn n n   ways. 1.1.2 The Multiplication Principle (AND) Suppose you want to buy an ice cream on a cone. There are 5 ice cream flavours to choose from and there are 3 types of cones. How many different choices do you have? No. of ways to choose a cone = 3 No. of ways to choose a flavour = 5 Total number of ways an ice cream AND a cone = 3 × 5 = 15 In general, if one operation can be done in n1 ways, and ( when this has been done ), the second operation can be performed in n2 ways and so on for k operations, then the k successive operations can be performed in 12 ... kn n n   ways. Cone 1. Waffle 2. Biscuit 3. Cup Flavour 1. Chocolate 2. Vanilla 3. Strawberry 4. Oreo 5. Mint

SAJC 2021 JC1 H1 Mathematics Chapter 1: Permutation and Combinations 2 Example 1 Mary and John go to a restaurant that offers 5 types of pizza and 6 types of salads. Mary would like to have either a pizza or a salad. John would like to have a pizza and a salad. How many choices do Mary and John have respectively? Solution: Mary John Case 1 Chooses pizza: 5 ways Case 2 Chooses salad: + 6 ways No. of choices = 11 Stage 1 pizza Stage 2 salad No. of choices = 5 ways x 6 ways = 30 Exercise 1 A bookshelf holds 6 different English books, 8 different Malay books and 10 different Chinese books. (a) How many ways are there to select 1 book in any of the languages? (b) How many ways are there to select 3 books, 1 in each language? (c) How many ways are there to select 2 books in 2 languages? [Ans: (a) 24; (b) 480; (c) 188] Solution (a) No. of ways = 6 + 8 + 10 = 24 (b) No. of ways = 6 8 10 = 480 (c) 3 cases: The 2 books can be English and Malay or English and Chinese or Malay and Chinese. No. of ways = 6 8 6 10 8 10     = 188

SAJC 2021 JC1 H1 Mathematics Chapter 1: Permutation and Combinations 3 1.2 PERMUTATIONS A permutation is an arrangement of objects where the order is important. Example 1. Permutations (or arrangements) that can be formed with the letters A, B, C by taking three at a time. ____ ____ ____ The first slot can have 3 choices (A, B or C ) .The second slot can have 2 choices. The third slot can have 1 choice. No of ways to arrange A, B, C = 3 × 2 × 1 = 3! = 6 ways ABC BAC CAB ACB BCA CBA 2. Permutations (or arrangements) that can be formed with the letters A, B, C by taking two at a time. ____ ____ In this case, the first slot can have 3 choices (A, B or C ) .The second slot can have 2 choices. = {AB, AC, BA, BC, CA, CB} = 3 × 2 = 6 ways Example 2 Without replacement (i.e. no repetition) How many 5-letter code-words can be formed from the letters of the word MATRICES? _____ _____ _____ _____ _____ Number of 5-letter code-words = 8 7 6 5 4 6720     Unless otherwise stated, a permutation of objects refers to one without replacement. The symbol Pn r is used to represent the number of permutations of r objects out of a total of n objects. Thus, the answer in Example 2(a) is 8 5P . There is a GC command for it: Press » and press ~ to PROB, Press ‘2’.

SAJC 2021 JC1 H1 Mathematics Chapter 1: Permutation and Combinations 4 In particular, the number of permutations of n distinct objects taken all at a time is P ( 1)...(2)(1) !.n n n n n   E.g. 8 8P 8! The number of permutations of r distinct objects out of n distinct objects is !P ( 1)...( 1) ( )! n r nn n n r nr      E.g. 8 5 8!P 8 7 6 5 4 3!      Example 3 (a) How many 4-digit numbers greater than 2000 can be formed from {0,1,2,3,4} if no repetition is allowed? (b) How many of these numbers from (a) will be divisible by 5 if no repetition is allowed? (c) How many of these numbers from (a) are even if no repetition is allowed? Solution: (a) ___ ___ ___ ___ 4 digits The first slot can only be taken by the numbers 2,3 or 4. The subsequent slots can only be taken by the remaining numbers and arranged in 4 3P ways. No. of numbers = 4 33 P 72 (b) ___ ____ ____ __0__ (the 4th slot has to be 0) The first slot can take 2, 3 or 4, the last slot has to take 0, the remaining 3 numbers will be used to fill the remaining two slots. No. of numbers = 3 23 P 1 18   (c) Case 1: ____ ____ ____ ____ 3 digits 2,4 0,2,4

SAJC 2021 JC1 H1 Mathematics Chapter 1: Permutation and Combinations 5 The first slot is taken by an even number 2 or 4. (2 ways) The last slot has to be taken by 0 or the other even number not taken by first slot. (2 ways) No. of numbers = 3 22 P 2 24   Case 2: _3__ ____ ____ ____ 3 digits The first slot is taken by an odd number 3. (1 way) The last slot has to be taken by either 0, 2, or 4. (3 ways) No. of numbers = 3 21 P 3 18   Total no. of numbers = 24 + 18 = 42 Exercise 2 1(i) How many 3-digit numbers can be formed from the digits { 4, 5, 6, 7, 8 } if repetitions of number are not allowed? (ii) Find the number of ways that a 3 -digit number can be formed from the digits {4, 5, 6, 7, 8 } if repetitions of number are not allowed and that it (a) is an odd number, (b) is an even number, (c) contains the digit 4 at most once, (d) contains the digit 8 twice. [Ans : (i) 60; (ii)(a) 24 ; (ii)(b) 36 ; (ii)(c) 60 ; (ii)(d) 12] Solution: (i) No. of ways = 5 x 4 x 3 = 60 or 5 3P (ii) (a)No. of odd numbers = 4