NO ONE CAN PREDICT TO WHAT HEIGHTS YOU CAN SOAR EVEN YOU WILL NOT KNOW UNTIL YOU SPREAD YOUR WINGS.

NO ONE CAN PREDICT TO WHAT HEIGHTS YOU CAN SOAREVEN YOU WILL NOT KNOW UNTIL YOU SPREAD YOUR WINGS

Chapter 19

PERMUTATIONS AND

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

City C

City B

Total number of ways = 3 x 4 = 12

City A

Solution:

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 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(r-s) Principle

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.

Multiplication Principle

Certainly, we can extend the Multiplication Principle to handle a task requiring more than 2 stages:

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

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 ?

- TWO BASIC PRINCIPLES-

Addition PrincipleExample 2:

Solution: Task: Travel to Tokyo from Singapore

Task by air – 4 ways

Task by sea – 3 ways

Total number of ways = 4 + 3 = 7

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

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

Example 2a:

Most questions may be a combination of the two different principles as shown below:

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-

Addition Principle

- TWO BASIC PRINCIPLES-

Ex 2a: Solution

ArtScience

ArtMaths

Science

Maths

2 books, 2 different subjects

= 3 x 5 = 15

= 5 x 2 = 10

= 3 x 2 = 6

Total = 31

(A different presentation)

PERMUTATIONS

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.

DEFINITION

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

Illustration

S = { } = set of 5 objects

Permutations( order within group is considered)

Of 4 objects

different

permutations

of 4objects

Of 3 objects

different

permutations

of 3objects

- 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)

=  

=

rnP

r)!(n

n!

12...)1)((

12...)1)()(1(...)2)(1(

rnrn

rnrnrnnnn

- PERMUTATIONS-

Interpretation of rnP

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

nnP

- PERMUTATIONS-

Rules Of Permutations

nn

all

different objec

L1: No. of arrangementsof n

in a row / line Ps nt !

nr

r

differ

L2 : No. of a

ent object

rrangementsof out of n

in a row / l nes i P

In how many ways can the letters A, B, C be arranged ?

Solution:

No. of ways = = 3! = 6

- PERMUTATIONS-

Example 3

33P

There are 10 vacant seats in a bus. In how many ways can 2 people seat themselves ?

- PERMUTATIONS-

Example 4

Solution:

No. of ways = 10P2

= 10 x 9

= 90

Find the number of ways of filling 5 spaces by selecting any 5 of 8 different books.

- PERMUTATIONS-

Example 5

Solution:

No. of ways = 8P5

= 8 x 7 x 6 x 5 x 4

= 6720

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)

- CONDITIONAL PERMUTATIONS-

Example 6

Arrange using 4 letters ( minus E or I)

E,I

No of arrangements beginning with a vowel

= 2 x 4! = 48

- PERMUTATIONS-

Rules Of Permutations

nn

all

different objec

L1: No. of arrangementsof n

in a row / line Ps nt !

nr

r

differ

L2 : No. of a

ent object

rrangementsof out of n

in a row / l nes i P

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

- CONDITIONAL PERMUTATIONS-

Example 7

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

Be objective

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

- CONDITIONAL PERMUTATIONS-

Example 7

Solution:No. of ways to arrange all 6 letters = 6! = 720No. of arrangements in which the vowels are together = 144Therefore, the no. of arrangements in which the vowels are not all together = 720 – 144 = 576

In how many ways can the letters of the word ‘ORANGE’ be arranged with the restriction that (iii) the 3 vowels are all separated?

- CONDITIONAL PERMUTATIONS-

Example 7

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:

C C C

- CONDITIONAL PERMUTATIONS-

Example 7

C being the position of a consonant and the position of a vowel.

Thus the number of ways of arranging ‘ORANGE’ so

that the 3 vowels are separated = 3! X = 14443P

C C C

The 3 C’s can be arranged in 3! ways; and we can use 4 positions to put in the 3 vowels in ways.4

3P

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

- CONDITIONAL PERMUTATIONS-

Example 8

(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

2,4,6

(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

- CONDITIONAL PERMUTATIONS-

Example 8

1 5

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

Possible selections or combinations of 3 objects drawn from the set { A, B, C, D } are

43C

Illustration

{ 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! 4

2C

COMBINATIONS

Possible selections or combinations of 2 objects drawn from the set { A, B, C, D,E } are

52C

An exercise

{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

Possible selections or combinations of 3 objects drawn from the set { A, B, C, D,E } are

53C

An exercise

{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

Possible selections or combinations of 6 nos. drawn from the set { 1,2,3,4,…,45 } are

456C

Illustration

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

rnC

!

)1)...(2)(1(

)!(!

!

r

rnnnn

rnr

n

!r

Prn

rnC =

=

In how many ways can a committee of 3 be chosen form 10 persons ?

Solution:

No. of ways = 10C3 = 120

- COMBINATIONS-

Example 9

In how many ways can 10 boys be divided into 2 groups (i) of 6 and 4 boys (unequal sizes)

Solution:

- COMBINATIONS-

Example 10

Select

First group of 6

Select

Second group of 4

No. of ways of forming groups = 10C6 X 4C4

= 210

In how many ways can 10 boys be divided into 2 groups (ii) of 5 and 5 boys (equal sizes)

Solution:

- COMBINATIONS-

Example 10

Select

First group of 5

Select

Second group of 5

No. of ways of forming groups = (10C5 X 5C5)/2! = 126

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

- COMBINATIONS-

Example 11

Solution:

a) 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 = 6No. of committees with at least 1 junior = (total no. of committees) – (no. of committees with no junior)= 252 – 6 = 246

the end