Paing Education Centre(Grade-12)
Sayar Pyae Sone
Chapter-5
Permutation and Combination
✓ Permutation is arrangement of items in which order matters. Eg- Arranging numbers , alphabets,
people , etc…
✓ Combination is the selection of items in which order does not matter. Eg – Selection of subjects,
menu, clothes, etc…
✓ In this chapter, the concepts of “permutation” and “combination” are described in accordance
with some counting principles.
5.1 Counting Principle
The Multiplication Principle (Successive tasks)
Suppose a task can be performed in m ways , and no matter how the task has been performed
another task can then be performed in n ways. Then the number of ways to perform these two tasks in
succession is mn (ways).
Number of ways first task can be perform = m ways
လုပ်ဆ
Number of ways another task can be perform = n ways
က်တုက်လုပဆ
်
Total number of ways for two tasks = mn
ောင်ချက်မျောျားသည်
ောင်နုင်လျှင်
Multiplication Principle ကုသုျားသည်။
Example Suppose a coffee shop offers 2 choices (Tea and Coffee) and 3 choices of snacks (Cake,
Doughnut and Sandwich). How many different choices are possible to have a drink and a snack?
Drink
Tea
Coffee
Snack
Possible Choices
Cake
( Tea , Cake )
Doughnut
( Tea , Doughnut)
Sandwich
( Tea , Sandwich)
Cake
( Coffee , Cake )
Doughnut
( Coffee , Doughnut)
Sandwich
( Coffee , Sandwich)
There are 6 different choices to have a drink and a snack.
Number of possible choices = (number of ways for 1st item) × ( number of ways for 2nd item)
Paing Education Centre(Grade-12)
Sayar Pyae Sone
Example(1) A list of 5 topics is given in an essay contest. A Student must select one of the topics on
which to write a short essay, and then select a different topic from the list for a long essay. How many
ways can the topics be chosen for the two ways.
Solution;
Number of ways for short essay = 5
Number of ways for long essay = 4
The number of possible choices = 5 × 4 = 20
Example(2) If you have 5 different shirts and 7 different hats, how many ways the total number of
different selections a shirt and a hat ?
Solution; Number of ways for shirt = 5
Number of ways for hat = 7
the total number of different selections a shirt and a hat = 5 × 7 = 35
Example(3) In a class, there are 16 boys and 9 girls. The Teacher wants to selected a boy for president
and a girl for vice-president for class. How many ways can be the teacher make for president and
vice- president?
Solution; Number of ways for president = 16
Number of ways for vice- president = 9
The number of ways can be the teacher make for president and vice- president = 16 × 9 = 144
Example(1) / Page (70)
Suppose that there are 6 roads between town A and the town B and that 4 roads between town B and
town C . Find the number of ways a person can drive from A to C by passing through B?
Solution;
Number of ways from A to B = 6
Number of ways from B to C = 4
The number of ways to drive from A to C by passing through B = 6 × 4 = 24
******
Paing Education Centre(Grade-12)
Sayar Pyae Sone
Example(2) / Page (70)
There are four blood types, namely A, B , AB and O. Blood can also be RH(+ve) or RH(-ve). A b lood donor
can be classified as either male or female. How many different possible ways can a donor have his or her
blood labeled?
Solution;
Number of blood types = 4
Number of blood types upon RH factor = 2
Number of types of donor = 2
The number of ways to label = 4 × 2 × 2 = 16
Example(3) / Page (70)
There are 3 pictures nails 0n a wall. If there are 5 different pictures and each nail can hold only one
picture, in how many different ways can the pictures be hung on all the nail?
Solution; Number of ways for the 1stnail = 5
Number of ways for the 2nd nail = 4
Number of ways for the 3rd nail = 3
The number of different ways to hung the pictures in all nails = 5 × 4 × 3 = 6
Addition Example (Page-71)
A manufacturer makes shirts in 5 different sizes S, M, L , XL and XXL. Shirts having sizes S, M and L are
made in 4 different designs, while those having sizes XL and XXL are in 3 different designs. If each design
has 5 different color schemes, how many different types of shirts are possible to make?
Solution;
(1) Shirts have sizes S, M or L.
Sizes
Design
Color
3
4
5
2
3
5
For the first case, the number of types = 3 × 4 × 5 = 60
For the second case, the number of types = 2 × 3 × 5 = 30
The total number of types = 60 + 30 = 90
*************
(2) Shirts have sizes XL or XXL.
Paing Education Centre(Grade-12)
Sayar Pyae Sone
The Addition Principle (Disjoint Case)
If a event can be performed in m ways and another event which is independent to the first, can be
performed n ways. The either of the two events can be performed in m + n ways.
➢ Suppose there are k pairwise disjoint cases, into which the elements in a counting problem fall.
➢ If there are n1 elements in the first case, n2 elements in the second case, n3 elements in the third
case, … , and nk elements in the kth case, then the number of elements in the problem is
n 1 + n 2 + n3 + … + nk
ဖြစ်ရပ်မျောျား တစ်ခုနှင်တစ်ခု မသက်
ုင် (တစ်ခုပပျားမှတစ်ခုဖြစ်) လျှင် Addition Principle ကုသုျားသည်။
Example(1) In a class, there are 16 boys and 9 girls. The Teacher wants to selected either a boy or a girl
as a class representative. How many ways can be the teacher make the selection?
Solution; There are 16 ways out of 16 boys.
There are 9 ways out of 9 girls.
The number of ways can the teacher make the selection = 16 + 9 = 25
Example(2) An examination has ten questions in section A and four questions in section B. How many
different ways are there to choose a question from either section A or section B?
Solution; There are 10 ways in section A.
There are 4 ways in section B.
Required number of ways = 10 + 4 = 14
Example(4) / Page (71)
How many different numbers can be formed using the digits 3, 5, 6, 8 and 9 in such a way that the
numbers contain two or three digits without any repetition?
Solution; The two-digit numbers can be formed in 5 × 4 = 20 ways
The three-digit numbers can be formed in 5 × 4 × 3 = 60 ways
There are 20 + 60 = 80 different number.
Example(5) / Page (72)
How many integers, having digit 5 only once, are there between 0 and 100 ?
Solution; The integers must be one of the following:
(i) One-digit integer, namely 5
(ii) Two-digit integers having 5 as tens’ digit (ones’ digit may be anyone except 5)
(iii) Two-digit integers having 5 as ones’ digit (tens’ digit may be anyone except 0 and 5)
For the first case, the number of integers = 1
For the second case, the number of integers = 1 × 9 = 9
For the first case, the number of integers = 8 × 1 = 8
The required number of integers = 1 + 9 + 8 = 18.
Factorial Notation
There are 7 pictures nails on the wall. If there are 7 different pictures and each nail can hold only one
picture, in how many different ways can the pictures be hung on all the nail?
1st nail
2nd nail
3rd nail
4thnail
5th nail
6thnail
7th nail
7
6
5
4
3
2
1
By the multiplication principle, the pictures can be hung in 7 × 6 × 5 × 4 × 3 × 2 × 1 different ways.
That is the product of the first seven positive integers and can be expressed as 𝟕! .
Generally, the product of the first n positive integers and can be expressed as 𝒏!
If n is a positive integer, 𝑛! = n (n - 1) (n - 2) … 1
Note; 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 8 × 𝟕! = 8 × (8 – 1 )!
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 7 × 𝟔! = 7 × (7 – 1) ! and so on.
Generally for positive integer n ,
n! = n (n - 1) (n - 2) … 1
n! = n (n - 1) !
n! = n (n - 1) (n - 2) !
1! = 1 ,
0! = 1
Proof ; 0! = 1
➢ n! = n (n - 1) !
1! = 1 (1 - 1) !
1 = 1 × 0!
/
So, 0! = 1
Example(6) / Page (72)
Evaluate: (a)
7!
6!+5!−4 !
(b)
4! . 3!
.
4!
Solution;
(a)
(b)
7!
4! .3!
=
6!+5!−4!
4!
7. 6 . 5 . 4!
=
4! .3 .2 .1
= 35
6 . 5 . 4! + 5 . 4! − 4!
4!
=
( 30+5−1) . 4!
= 34
4!
Example(7) / Page (72)
13 . 12 . 11
in factorial form.
3 . 2
13 . 12 . 11 13 . 12 . 11 . 10!
13!
Solution;
=
=
( 3 . 2 . 1 ) . 10!
3 . 2
3! . 10!
Express
___________________________________________________________________
Exercise(5.1)
1. A store sells men’s wear. It has 6 kinds of shirts, 4 kinds of pants and 3 kinds of coats. If a man wants
to buy a shirt, a pant and a coat, in how many ways can this be done? (Assume that any choice meets
his requirement.)
Solution;
Want to buy
Shirt
Pant
Coat
Number of ways
6
4
3
Possible number of ways = 6 × 4 × 3 = 72
There are 72 ways a man can buy a shirt, a plant and a coat.
2. A television news director wishes to use three of the 7 news stories on an evening show. How many
possible ways can be program be set up, if the three stories are to be classified as the lead story, the
second story and closing story?
Solution;
Type of story
Lead story
Second story
Closing story
Number of ways
7
6
5
Possible number of ways = 7 × 6 × 5 = 210
There are 210 ways can be set up the program.