1. Ronald Hui Tak Sun Secondary School
HKDSE Mathematics
Ronald Hui
Tak Sun Secondary School

2. Homework SHW7-B1: Sam L SHW7-P1: Sam L RE8: Sam L
EHHW1: Tashi, Kelvin, Sam L, Pako
EHHW2: Tashi, Kelvin, Sam L, Pako
Ronald HUI

3. Homework SHW9-B1: Sam L SHW9-C1: Kelvin (RD), Sam L
SHW9-D1: Tashi, Matthew (RD), Ken (RD), Kelvin, Yan Tin, Walter
SHW9-E1: Tashi, Kelvin
SHW9-R1: Tashi, Kelvin, Sam L
SHW9-P1: Tashi, Sam L
RE9: Sam L, Pako
Ronald HUI

4. Basic Principles of Counting

5. In how many ways can I choose a pet to buy?
Addition Rule of Counting
There are 7 cats and 5 dogs in a pet shop.
In how many ways can I choose a pet to buy?

6. Addition Rule of Counting
Method 1: There are 7 ways of choosing a cat.
Method 2: There are 5 ways of choosing a dog.
∴ The total number of ways of choosing a pet

7. You have applied the addition rule of counting.
In general,
suppose a task can be performed by k methods, in which there are
n1 ways to perform method 1,
n2 ways to perform method 2, … ,
nk ways to perform method k,
then there are in total n1 + n2 + … + nk ways to perform the task.

8. If a task can be performed using two methods which have common ways, can we still apply the addition rule of counting?
In this case, we need to subtract the common ways when we apply the rule to avoid double counting. Let’s see an example.

9. How many different choices are there if a sticker is chosen?
Suppose Mary has 6 different stickers, Ada has 4 different stickers and they have 2 stickers in common.
How many different choices are there if a sticker is chosen? E F C D
Mary’s stickers H G
Ada’s stickers B A B A
The required number of ways

10. In general, we have the following rule.
If a task can be performed using either method 1 with n1 ways
or method 2 with n2 ways,
and there are m ways in common,
then the number of ways to perform the task is n1 + n2 – m.
Method 1 n1
Method 2 n2 m

11. Follow-up question
A number is chosen from 1 to 60 inclusive. How many ways of selecting a number that
(a) is either a multiple of 6 or 11?
(b) is either a multiple of 3 or 11?
(a)
By the addition rule of counting,
the required number of ways =
There are 10 multiples of 6: 6, 12, 18, ... , 60.
There are 5 multiples of 11: 11, 22, 33, 44, 55.
There are no common multiples of 6 and 11.
= 15

12. Follow-up question
A number is chosen from 1 to 60 inclusive. How many ways of selecting a number that
(a) is either a multiple of 6 or 11?
(b) is either a multiple of 3 or 11?
(b)
By the addition rule of counting,
the required number of ways
= – 1
There are 20 multiples of 3: 3, 6, 9, ... , 33, ... , 60.
There are 5 multiples of 11: 11, 22, 33, 44, 55.
There is 1 common multiple of 3 and 11: 33.
= 24

13. Multiplication Rule of Counting
A restaurant offers 3 kinds of bread and 2 kinds of drinks.
I want to have a bread and a drink for my breakfast. How many choices do I have?

14. ∴ Total number of choices
1st step
2nd step
Possible outcomes
∴ Total number of choices
= 3  2
Number of kinds of bread = 3
Number of kinds of drinks = 2
= 6

15. In general, we can use the multiplication rule of counting to solve this kind of problem.
Suppose a task involves a sequence of k steps. If there are
n1 ways to perform step 1,
n2 ways to perform step 2, … ,
nk ways to perform step k,
then the number of ways to perform the task is n1  n2  …  nk.

16. Example:
There are 3 different pencils, 4 different rulers and 6 different books in a bag. Find the number of ways of choosing a pencil, a ruler and a book from the bag.
The required number of ways
= 3  4  6
= 72
Some problems may involve both the addition rule and the multiplication rule of counting. Let’s see an example.

17. There are 8 Japanese courses, 6 French courses and 5 German courses
There are 8 Japanese courses, 6 French courses and 5 German courses. If Victor wants to select two courses of different languages, how many choices are there?
There are 3 possible types:
Type Japanese course + French course
Type Japanese course + German course
Type French course German course

18. There are 8 Japanese courses, 6 French courses and 5 German courses
There are 8 Japanese courses, 6 French courses and 5 German courses. If Victor wants to select two courses of different languages, how many choices are there?
Type 1
Number of choices for a Japanese course and a French course
= 8  6 = 48
By multiplication rule of counting
Type 2
Number of choices for a Japanese course and a German course
= 8  5 = 40
By multiplication rule of counting
Type 3
Number of choices for a French course and a German course
= 6  5 = 30
By multiplication rule of counting

19. By addition rule of counting
There are 8 Japanese courses, 6 French courses and 5 German courses. If Victor wants to select two courses of different languages, how many choices are there?
∴ The required number of choices =
= 118
By addition rule of counting
Do you know why the addition rule of counting is used in the last step?

20. Follow-up question
There are 15 boys and 13 girls in both class A and class B. If a boy and a girl are selected from the same class to participate in an exchange program, how many choices are there?
By multiplication rule of counting,
number of choices of selecting a boy and a girl from class A
Similarly,
number of choices of selecting a boy and a girl from class B
By addition rule of counting,
the required number of choices

21. Permutation

22. Factorial Notation
We use the notation n! to denote the product of the first n positive integers from 1 to n.
 n! is read as n factorial.
For any positive integer n, .
Moreover, we define 0! = 1.

23. In fact, we can find the value of n! by calculator. For example,
keying in gives the value of 7! , which is 5040.

24. But we can also evaluate expressions involving factorial without calculator. For example,
Factorial notation is useful in finding the number of arrangements or permutations.

25. We may also arrange the pictures in this way.
Permutation
Consider 2 different pictures on a wall.
Picture A
Picture B
Picture B
Picture A
We may also arrange the pictures in this way.

26. In daily life, we often come across situations of arranging objects.
Permutation
Consider 2 different pictures on a wall.
Picture A
Picture B
Picture B
Picture A
In daily life, we often come across situations of arranging objects.

27. An arrangement of a certain number of objects in a definite order is called a permutation.
Let us study how to find the number of permutations of several distinct objects.

28. Suppose we want to arrange the following 3 soldiers in a row
Suppose we want to arrange the following 3 soldiers in a row. How many different arrangements are there? A B C
Number of choices
1st
2nd
3rd

29. Suppose we want to arrange the following 3 soldiers in a row
Suppose we want to arrange the following 3 soldiers in a row. How many different arrangements are there? A B C
Number of choices
1st
2nd
3rd 3 
There are 3 ways to choose the first soldier.
Step 1

30. Suppose we want to arrange the following 3 soldiers in a row
Suppose we want to arrange the following 3 soldiers in a row. How many different arrangements are there? A B C B C A
Number of choices
1st
2nd
3rd 3 2 
There are 2 ways to choose the second soldier.
Step 2

31. Suppose we want to arrange the following 3 soldiers in a row
Suppose we want to arrange the following 3 soldiers in a row. How many different arrangements are there? A B C C B A
____ B C A
Number of choices
1st
2nd
3rd 3 2 1 
There are 1 way to choose the third soldier.
Step 3

32. Suppose we want to arrange the following 3 soldiers in a row
Suppose we want to arrange the following 3 soldiers in a row. How many different arrangements are there? A B C C B A
____ B C A
ABC
ACBBAC
BCA
CAB
CBA
6 possible outcomes
Number of choices
1st
2nd
3rd 3 2 1
By the multiplication rule of counting,
number of permutations

33. Now we want to arrange the following 4 soldiers in a row
Now we want to arrange the following 4 soldiers in a row. How many different arrangements are there?
Number of choices
1st
2nd
3rd
4th 4 
There are 4 ways to choose the first soldier.
Step 1

34. Now we want to arrange the following 4 soldiers in a row
Now we want to arrange the following 4 soldiers in a row. How many different arrangements are there?
Number of choices
1st
2nd
3rd
4th 4 3 
There are 3 ways to choose the second soldier.
Step 2

35. Now we want to arrange the following 4 soldiers in a row
Now we want to arrange the following 4 soldiers in a row. How many different arrangements are there?
Number of choices
1st
2nd
3rd
4th 4 3 2 
There are 2 ways to choose the third soldier.
Step 3

36. Now we want to arrange the following 4 soldiers in a row
Now we want to arrange the following 4 soldiers in a row. How many different arrangements are there?
Number of choices
1st
2nd
3rd
4th 4 3 2 1 
There are 1 way to choose the fourth soldier.
Step 4

37. Now we want to arrange the following 4 soldiers in a row
Now we want to arrange the following 4 soldiers in a row. How many different arrangements are there?
Number of choices
1st
2nd
3rd
4th 4 3 2 1
By the multiplication rule of counting,
number of permutations

38. In some situations, not all of the n distinct objects are arranged.
By similar arguments, we have:
The number of permutations of n distinct objects without repetition is n!.
In some situations, not all of the n distinct objects are arranged.

39. By similar arguments, we have:
The number of permutations of n distinct objects without repetition is n!.
Let us consider the case when we only arrange some of the objects but not all of them.

40. If 2 soldiers are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd 4 
There are 4 ways to choose the first soldier.
Step 1

41. If 2 soldiers are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd 4 3 
There are 3 ways to choose the second soldier.
Step 2

42. If 2 soldiers are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd 4 3
By the multiplication rule of counting,
number of permutations

43. If 3 soldiers (instead of 2) are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd
3rd 4 
There are 4 ways to choose the first soldier.
Step 1

44. If 3 soldiers (instead of 2) are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd
3rd 4 3 
There are 3 ways to choose the second soldier.
Step 2

45. If 3 soldiers (instead of 2) are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd
3rd 4 3 2 
There are 2 ways to choose the third soldier.
Step 3

46. If 3 soldiers (instead of 2) are selected and arranged in a row, how many different arrangements are there?
Number of choices
1st
2nd
3rd 4 3 2
By the multiplication rule of counting,
number of permutations

47. By similar arguments, we have:
The number of permutations of n distinct objects taken r at a time without repetition,
denoted by , . is

48. By similar arguments, we have:
The number of permutations of n distinct objects taken r at a time without repetition,
denoted by , . is
0! = 1
In particular,

49. Example:
The number of permutations of 5 distinct balls taken 2 at a time without repetition is
P = 5  4 = 20. 2 5
One possible permutation:
1st
2nd

50. Example:
The number of permutations of 6 distinct balls taken 2 at a time without repetition is
P = 6  5 = 30. 2 6
One possible permutation:
1st
2nd

51. Example:
The number of permutations of 6 distinct balls taken 3 at a time without repetition is
P = 6  5  4 = 120. 3 6
One possible permutation:
1st
2nd
3rd

52. In fact, we can find the value of by calculator. For example,
keying in gives the value of , which is 56.
P 8 2

53. Follow-up question
(a) means the number of permutations of ___ distinct objects taken ___ at a time without repetition.
(b) Evaluate the following expressions.
(i) = _____ (ii) = _______
P 8 4 8 4
210
P 7 3
P 14 5

54. Let us see how to use the above results to solve permutation problems.
Recall:
1. The number of permutations of n distinct objects without repetition is n!.
2. The number of permutations of n distinct objects taken r at a time without repetition,
denoted by ,
is n  (n  1)  (n  2)  ...  (n  r + 1) =
Let us see how to use the above results to solve permutation problems.

55. There are 7 distinct digits.
Example:
There are 7 distinct numbers 1, 3, 4, 5, 7, 8 and 9.
(a) How many 7-digit numbers can be formed by using the above 7 numbers without repetition?
There are 7 distinct digits.
(a)
Number of 7-digit numbers formed

56. There are 7 distinct digits.
Example:
There are 7 distinct numbers 1, 3, 4, 5, 7, 8 and 9.
(b) How many 3-digit numbers can be formed by using the above 7 numbers without repetition?
There are 7 distinct digits.
(b)
Number of 3-digit numbers formed

57. Example: There are 7 distinct numbers 1, 3, 4, 5, 7, 8 and 9.
(c) How many 3-digit numbers formed in (b) are divisible by 5?
(c)
For a 3-digit number formed in (b) to be divisible by 5,
its last digit must be 5. 5
Choose 2 numbers from the remaining 6 numbers and arrange them in order.

58. Example: There are 7 distinct numbers 1, 3, 4, 5, 7, 8 and 9.
(c) How many 3-digit numbers formed in (b) are divisible by 5?
(c)
For a 3-digit number formed in (b) to be divisible by 5,
its last digit must be 5.
∴ Number of choices for 1st digit and 2nd digit 5
∴ The required number is 30.

59. O G E N R A Follow-up question 6 distinct letters are shown below.
(a) How many 6-letter strings can be formed by using the above 6 letters without repetition?
(b) How many 4-letter strings can be formed by
using the above 6 letters without repetition? G O E R N A
(a) Number of 6-letter strings formed =

60. O G E N R A Follow-up question 6 distinct letters are shown below.
(a) How many 6-letter strings can be formed by using the above 6 letters without repetition?
(b) How many 4-letter strings can be formed by
using the above 6 letters without repetition? G O E R N A
(b) Number of 4-letter strings formed =

61. Example:
4 male and 5 female models are selected from 6 male and 8 female models to stand on a stage. In how many ways can they be arranged in a row if models of the same sex must stand next to each other?
Suppose the male models stand on the left.
4 male models
5 female models
Number of ways of choosing 4 male models from the 6 male models and arrange them in order
Number of ways of choosing 5 female models from the 8 female models and arrange them in order

62. By multiplication rule of counting
Example:
4 male and 5 female models are selected from 6 male and 8 female models to stand on a stage. In how many ways can they be arranged in a row if models of the same sex must stand next to each other?
Suppose the male models stand on the left.
4 male models
5 female models
∴ Number of ways of arranging the 4 male models on the left and the 5 female models on the right
By multiplication rule of counting

63. 4 male models
5 female models
4 male models
5 female models
Number of ways of arranging the 4 male models on the left
and the 5 female models on the right
5 female models
4 male models
Number of ways of arranging the 5 female models on the left and the 4 male models on the right
The number of ways of arranging the male models on the right is the same as that of arranging the male models on the left.

64. 4 male models
5 female models
4 male models
5 female models
Number of ways of arranging the 4 male models on the left
and the 5 female models on the right
5 female models
4 male models
Number of ways of arranging the 5 female models on the left and the 4 male models on the right
∴ The required number of arrangements

65. Follow-up question
2 boys and 3 girls are selected from 5 boys and 7 girls and stand in a queue. In how many ways can they be arranged if all the girls must stand in front of the boys?
3 girls
2 boys
Number of ways of choosing 3 girls from the 7 girls and arrange them in a queue
Number of ways of choosing 2 boys from the 5 boys and arrange them in a queue
∴ The required number of ways