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

2. Chapter 19 PERMUTATIONS AND
COMBINATIONS

3. 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 :

4. 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?

5. 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 ?

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

7. 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.

8. 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.

9. 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.

10. - 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.

11. - 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 = = 7

12. - 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 .

13. - 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?

14. - 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 = = 31

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

16. 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.

17. 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.

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

19. - 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) =

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

21. - PERMUTATIONS- Rules Of Permutations

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

23. - 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

24. - 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

25. - 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

26. - 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

27. - PERMUTATIONS- Rules Of Permutations

28. - 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?

29. - 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

30. - 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

31. - 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:

32. - 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

33. - 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’.

34. - 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

35. - CONDITIONAL PERMUTATIONS- Example 81 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

36. 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.

37. COMBINATIONS
Consider the number of permutations of three different books A, B, C on a self:
ABC, ACB, BAC, BCA, CAB, CBA ! = 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}

38. 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!

39. 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!

40. 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!

41. 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!

42. 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 = =

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

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

45. - 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

46. - 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

47. - 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

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

49. - 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

50. - 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

51. the end