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Chapter 19 PERMUTATIONS AND COMBINATIONS

INTRODUCTION In our daily lives, we often need to enumerate “events” such as the arrangement of objects in a certain way, the partition of things under a certain condition, the distribution of items according to a certain specification and so on. In this topic, we attempt to formulate a general principle to help us to answer some counting problems like :

INTRODUCTION In how many ways can the numbers 0,1,2,3,4,5,6,7,8,9 be used to form a 4-digit number ? In how many ways can 4 representatives be chosen from a group of 10 students?

Multiplication Principle Example 1: There are 3 roads connecting city A and city B, and 4 roads connecting city B and city C. How many ways are there to travel from city A to city C via city B ?

Multiplication Principle Solution: City A City B City C Total number of ways = 3 x 4 = 12

Multiplication Principle In this example, we may regard travelling from A to C via B as a task, travelling from A to B as stage1, and from B to C as stage 2. stage1 and stage 2 are necessary or chained/linked for the task to be completed.

Multiplication(r-s) Principle Then we can develop the Multiplication Principle which states that if it requires two necessary stages to complete a task and if there are r ways and s ways to perform stage 1 and stage 2 respectively, then the number of ways to complete the task is r  s ways.

Multiplication Principle Certainly, we can extend the Multiplication Principle to handle a task requiring more than 2 stages: If a task/process consists of k linked steps/stages, of which the first can be done in n1 ways, the second in n2 ways,…, the r th step in nr ways, then the whole process can be done in n1  n2  …  nr  …  n k ways.

- TWO BASIC PRINCIPLES- Addition Principle If a task can be carried out through n possible distinct(mutually exclusive) processes(sub-tasks) and if the first process can be done in n1 ways, the second in n2 ways,…, the r th tasks in nr ways, then the task can be done in a total of n1 + n2 + … + nr + … + nk ways.

- TWO BASIC PRINCIPLES- Addition Principle Example 2: In how many ways can a man travel from Singapore to Tokyo if there are 4 airlines and 3 shipping lines operating between the 2 cities ? Solution: Task: Travel to Tokyo from Singapore Task by air – 4 ways Task by sea – 3 ways Total number of ways = 4 + 3 = 7

- TWO BASIC PRINCIPLES- Addition Principle Note: Since the man cannot travel by air and sea at the same time, we say that the two (approaches in carrying out the) tasks are mutually exclusive .

- TWO BASIC PRINCIPLES- Addition Principle Most questions may be a combination of the two different principles as shown below: Example 2a: In how many ways can we select 2 books of different subjects from among 5 distinct Science books, 3 distinct Mathematics books and 2 distinct Art books?

- TWO BASIC PRINCIPLES- Addition Principle Solution:There are three cases to consider:  Case 1: Select 1 Science & 1 Maths book. No. of ways = 5 x 3 = 15  Case 2: Select 1 Science & 1 Art book. No. of ways = 5 x 2 = 10  Case 3: Select 1 Maths & 1 Art book. No. of ways = 3 x 2 = 6   Hence, combining Case 1, 2 and 3, total no.of ways of selecting 2 books of different subjects = 15 + 10 + 6 = 31

- TWO BASIC PRINCIPLES- Ex 2a: Solution (A different presentation) Art Science Maths 2 books, 2 different subjects = 3 x 5 = 15 = 3 x 2 = 6 = 5 x 2 = 10 Total = 31

PERMUTATIONS DEFINITION Let S be a set of n objects. A permutation (arrangement) of r objects drawn from the set S( where r < n) is a sub-set of S containing r objects in which the order of the objects in the sub-set is taken into consideration.

PERMUTATIONS Are the following permutations? a) Arranging 3 different books on a shelf. b) Choose 4 books from a list of 10 titles to take away for reading. c) Seating 500 graduates in a row to receive their degree scrolls. d) Forming three-digit numbers using the integers 1, 2, 3, 4, 5. e) Select two students from a CG to attend a talk.

PERMUTATIONS S = { } = set of 5 objects Illustration Permutations( order within group is considered) Of 3 objects Of 4 objects

- PERMUTATIONS- Interpretation of When r objects taken from n different objects are permuted (arranged), the number of possible permutations denoted by nPr , is given by   nPr = n(n - 1)(n - 2) … (n - r + 1) =  

- PERMUTATIONS- Interpretation of Special case : When all the n different objects are permuted, thenumber of permutations is = = n!

- PERMUTATIONS- Rules Of Permutations

- PERMUTATIONS- Example 3 In how many ways can the letters A, B, C be arranged ? Solution: No. of ways = = 3! = 6

- PERMUTATIONS- Example 4 There are 10 vacant seats in a bus. In how many ways can 2 people seat themselves ? Solution: No. of ways = 10P2 = 10 x 9 = 90

- PERMUTATIONS- Example 5 Find the number of ways of filling 5 spaces by selecting any 5 of 8 different books. Solution: No. of ways = 8P5 = 8 x 7 x 6 x 5 x 4 = 6720

- CONDITIONAL PERMUTATIONS- Example 6 How many arrangements of the letters of the word ‘BEGIN’ are there which start with a vowel ? Solution: The first letter must be either E or I ie 2 ways. The rest can be arranged in 4! ways. Therefore no. of arrangements = 2 x 4! = 48

- CONDITIONAL PERMUTATIONS- Example 6 How many arrangements of the letters of the word ‘BEGIN’ are there which start with a vowel ? Solution:( A different presentation) E,I Arrange using 4 letters ( minus E or I) No of arrangements beginning with a vowel = 2 x 4! = 48

- PERMUTATIONS- Rules Of Permutations

- CONDITIONAL PERMUTATIONS- Example 7 In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (i) the 3 vowels must come together (ii) the 3 vowels must not come together (iii) the 3 vowels are all separated?

- CONDITIONAL PERMUTATIONS- Example 7 In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (i) the 3 vowels must come together Be objective Solution: { O,A,E } R N G (i) Strategy: bundle the 3 vowels together as 1 unit but remember you will have to arrange the 3 vowels within this unit in 3! ways. No. of words in which the 3 vowels are together = 4! x 3! = 144

- CONDITIONAL PERMUTATIONS- Example 7 In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (ii) the 3 vowels must not all come together Solution: No. of ways to arrange all 6 letters = 6! = 720 No. of arrangements in which the vowels are together = 144 Therefore, the no. of arrangements in which the vowels are not all together = 720 – 144 = 576

- CONDITIONAL PERMUTATIONS- Example 7 In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (iii) the 3 vowels are all separated? Solution: Qn: How do we separate some objects? Ans: We use the other objects as separators! Thus we use the consonants to separate the 3 vowels as in the configuration below:

- CONDITIONAL PERMUTATIONS- Example 7 C being the position of a consonant and the position of a vowel. The 3 C’s can be arranged in 3! ways; and we can use 4 positions to put in the 3 vowels in ways. Thus the number of ways of arranging ‘ORANGE’ so that the 3 vowels are separated = 3! X = 144

- CONDITIONAL PERMUTATIONS- Example 8 Given six digits 1,2,4,5,6,7. Find the number of 6-digit numbers which can be formed using all 6 digits without repetition if i) there is no restriction; ii) the numbers are all even; iii) the numbers begins with ‘1’ and end with ‘5’.

- CONDITIONAL PERMUTATIONS- Example 8 Solution: (i) No. of ways = 6! = 720 2,4,6 (ii) For each no. formed to be even, the last digit must be occupied by 2, 4 or 6 i.e. there are 3 ways to fill the last digit. The first 5 digits can be filled in 5! ways. Therefore, the total no. of even nos. than can be formed = 3 x 5! = 360

- CONDITIONAL PERMUTATIONS- Example 8 1 5 (iii) The first digit(1) and the last digit(5) are fixed (no arrangement needed). The remaining 4 digits can be filled in 4! ways. Therefore, the total no. of such nos. formed = 1 x 4! = 24

COMBINATIONS A Combination (Selection) of a given number of articles is a sub-set of articles selected from those given where the order of the articles in the subset is not taken into consideration. Examples of combinations a) Pick a set of three integers from the numbers 1, 2, 3, 4, 5. b) Choose a committee of 3 persons from a group of 10 people. c) From a box of 200 lucky draw tickets, 3 to be drawn as consolation prize.

COMBINATIONS Consider the number of permutations of three different books A, B, C on a self: ABC, ACB, BAC, BCA, CAB, CBA ----- 3! = 6 ways If order is not important, then there is only one way to choose all the three different books, i.e. simply {A, B, C}

COMBINATIONS Illustration Possible selections or combinations of 3 objects drawn from the set { A, B, C, D } are { A, B, C } , { A, B, D }, { A, C , D},{ B, C , D}. No. of possible selections or combinations of 3 objects drawn from the set { A, B, C, D } = 4 = = 4P3 / 3!

COMBINATIONS Possible selections or combinations of 2 objects drawn from the set { A, B, C, D } are { A, B } , { A, C }, { A, D } , { B, C }, { B, D } , { C, D }. = 6 or = 4P2 / 2!

COMBINATIONS An exercise Possible selections or combinations of 2 objects drawn from the set { A, B, C, D,E } are {A, B} , {A, C} , {A, D} , {A, E} , {B, C} , {B, D} , {B, E} , {C, D} , {C, E} , {D, E} No. of possible selections or combinations of 2 objects drawn from the set {A, B, C, D, E } = 10 = = 5P2 / 2!

COMBINATIONS An exercise Possible selections or combinations of 3 objects drawn from the set { A, B, C, D,E } are {A, B, C} , {A, B, D} , {A, B, E} , {A, C, D} {A, C, E} , {A, D, E} , {B, C, D} , {B, C, E} {B, D, E} , {C, D, E} No. of possible selections or combinations of 2 objects drawn from the set {A, B, C, D, E } = 10 = = 5P3 / 3!

COMBINATIONS Illustration Possible selections or combinations of 6 nos. drawn from the set { 1,2,3,4,…,45 } are impossible to list out without the aid of a computer. No. of possible selections or combinations of 6 nos. drawn from the numbers 1 to 45 inclusive = = 8 145 060

COMBINATIONS The number of combinations of r objects taken from n unlike objects is , where =

- COMBINATIONS- Example 9 In how many ways can a committee of 3 be chosen form 10 persons ? Solution: No. of ways = 10C3 = 120

- COMBINATIONS- Example 10 In how many ways can 10 boys be divided into 2 groups (i) of 6 and 4 boys (unequal sizes) Solution: Select First group of 6 Select Second group of 4 No. of ways of forming groups = 10C6 X 4C4 = 210

- COMBINATIONS- Example 10 In how many ways can 10 boys be divided into 2 groups (ii) of 5 and 5 boys (equal sizes) Solution: Select First group of 5 Select Second group of 5 No. of ways of forming groups = (10C5 X 5C5)/2! = 126

- COMBINATIONS- Example 11 A committee of 5 members is to be selected from 6 seniors and 4 juniors. Find the number of ways in which this can be done if a) there are no restrictions, b) the committee has exactly 3 seniors, c) the committee has at least 1 junior

- COMBINATIONS- Example 11 Solution: If there are no restrictions, the 5 members can be selected from the 10 people in 10C5 = 252 ways

- COMBINATIONS- Example 11 Solution: b) If the committee has exactly 3 seniors, these seniors can be selected in 6C3 ways. The remaining 2 members must then be juniors which can be selected from the 4 juniors in 4C2 ways. By the multiplication principle, the no. of committees with exactly 3 seniors is 6C3  4C2 = 120

- COMBINATIONS- Example 12 Solution: c) No. of committees with no juniors = 4C0  6C5 = 6 No. of committees with at least 1 junior = (total no. of committees) – (no. of committees with no junior) = 252 – 6 = 246

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