G10-MathQ3-Week-2- Definition of Permutation

Let’s bow our head and pray to our God, Almighty.
Dear God, We thank you for giving us another life, We
thank you for another Beautiful Day. As we go
through our lessons today, may you make us
instruments to do good things. Please enlighten our
minds, Give us the strength to participate in our
subject today. Thank you for this opportunity to learn
and serve others, and help me to always remember
the Truth of Your Glory! We offer this prayer and all
praise and glory to you, God Almighty, Amen.
PRAYER:

At the end of the lesson, I should be able to:
a. evaluate factorial notation
OBJECTIVES:

FACTORIAL NOTATION
The factorial n! is defined for a natural
number n as
n! = n (n-1)(n-2)…2 ⦁ 1.
The product of 4 ⦁ 3 ⦁ 2 ⦁ 1 can be written
simply as 4! (read as “4 factorial”).

n EXPANDED FORM n!
0!
1!
2!
3!
4!
5!
6!
7!
8!
9!
10!
1
1
2 ⦁ 1
3 ⦁ 2 ⦁ 1
4 ⦁ 3 ⦁ 2 ⦁ 1
5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
1
1
2
6
24
120
720
5, 040
40, 320
362, 880
3, 628, 800
(0! Is defined to be 1 so that
certain theorems and formulas
can be stated concisely.)

1) 4! + 3!
Solution:
4! + 3! = (4⦁3⦁2⦁1 ) + (3⦁2⦁1)
= 24 + 6
= 30
Examples: Evaluate.
2) 7! - 5!
Solution:
7! - 5! = (7⦁6⦁5⦁4⦁3⦁2⦁1) + (5⦁4⦁3⦁2⦁1)
= 5, 040 - 120
= 4, 920

3) 4!2!
Solution:
4!2! = (4⦁3⦁2⦁1 ) (2⦁1)
= (24) (2)
= 48
Examples: Evaluate.
4)
6!
3!
Solution:
6!
3!
=
6⦁5⦁4⦁3⦁2⦁1
3⦁2⦁1
= 120
5)
7!
(7−2)!
Solution:
7!
5!
=
7⦁6⦁5⦁4⦁3⦁2⦁1
5⦁4⦁3⦁2⦁1
=
7⦁6⦁5!
5!
6)
5!
3!2!
Solution:
5!
3!2!
=
5⦁4⦁3!
3!2!
= 42
= 42
=
5⦁4
2⦁1 =
20
2
= 10

Determine whether each of the following equation is True
or False.
1. 5! = 5 ⦁ 4 ⦁ 3!
5⦁4⦁3⦁2⦁1 = 5⦁4⦁3⦁2⦁1
120 = 120
2. 3! + 3! = 6!
(3⦁2⦁1) + (3⦁2⦁1) = 6⦁5⦁4⦁3⦁2⦁1
6 + 6 = 720
12 ≠ 720
TRUE
FALSE

Determine whether each of the following equation is True
or False.
3. 3!5! = 8!
(3⦁2⦁1)(5⦁4⦁3⦁2⦁1) =8⦁7⦁6⦁5⦁4⦁3⦁2⦁1
(6)(120) = 40, 320
720 ≠ 40, 320
4.
8!
5!
= 8⦁7⦁6
8⦁7⦁6⦁5!
5!
= 8⦁7⦁6
8⦁7⦁6 = 8⦁7⦁6
336 = 336
FALSE
TRUE

A. Evaluate.
1) (15-4)!
2)
𝟓!
(𝟓−𝟐)!
3)
𝟏𝟎!
𝟓!𝟒!𝟑!
4)
𝟔!
𝟑 𝟔−𝟐 !
5)
𝟗!
(𝟑!)𝟐

A. Evaluate.
1) (15-4)! = 11!
= 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
= 39, 916, 800
2)
𝟓!
(𝟓−𝟐)!
=
𝟓!
𝟑!
3)
𝟏𝟎!
𝟓!𝟒!𝟑!
4)
𝟔!
𝟑 𝟔−𝟐 !

A. Evaluate.
1) (15-4)! = 11!
= 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
= 39, 916, 800
2)
𝟓!
(𝟓−𝟐)!
=
𝟓!
𝟑!
=
𝟓•𝟒•𝟑!
𝟑!
= 5 • 4
= 20

At the end of the lesson, I should be able to:
a. Illustrates permutation of objects.
b. Finds the permutation of n objects taken r at a
time.
c. Illustrates distinguishable permutations.
d. Illustrates circular permutations.
e. Solves problems involving permutations.
OBJECTIVES:

Permutation
A permutation is an arrangement of objects in which
order is important.
Examples:
possible choices you have in creating your g-cash
password
arranging 5 different books in a shelf
possible no. of plate no. can LTO create

Methods in Determining Permutations
1. Systematic Listing
2. Tree Diagram
3. Two Way Table
4. Fundamental Counting Principle

Suppose you secure your bike using a
combination lock. Later, you realized that
you forgot the 4- digit code. You only
remembered that the code contains the
digits 2, 3, 4, and 7. List all the possible
codes out of the given digits. Determine the
number of permutations made.

Suppose you secure your bike using a combination lock.
Later, you realized that you forgot the 4- digit code. You
only remembered that the code contains the digits 2, 3, 4,
and 7. List all the possible codes out of the given digits.
Determine the number of permutations made.
2, 3, 4, 7
2, 3, 7, 4
2, 7, 3, 4
2, 7, 4, 3
2, 4, 7, 3
2, 4, 3, 7
3, 2, 4, 7
3, 2, 7, 4
3, 7, 2, 4
3, 7, 4, 2
3, 4, 7, 2
3, 4, 2, 7
4, 3, 2, 7
4, 3, 7, 2
4, 7, 3, 2
4, 7, 2, 3
4, 2, 7, 3
4, 2, 3, 7
7, 3, 2, 4
7, 3, 4, 2
7, 4, 3, 2
7, 4, 2, 3
7, 2, 4, 3
7, 2, 3, 4
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

To check our answer let’s use FCP
4 3 2
× × =
24 possible ways
× 1
2, 3, 4, and 7.
Given: 4 digit combination lock
𝟐
𝟒
𝟑
𝟕
𝟒
𝟑
𝟕
𝟒
𝟕 𝟕

A couple is planning to have 4 kids.
On any birth the children could be a
son or a daughter. Use S to represent
a son and D to represent a daughter.

S
S
D
𝟏. (𝑺, 𝑺, 𝑺, 𝑺)
A couple is planning to have 4 kids. On any birth the children could
be a son or a daughter. Use S to represent a son and D to represent a
daughter.
S
D
S
D
S
D
S
D
S
D
S
D
2. (𝑺, 𝑺, 𝑺, 𝑫)
3. (𝑺, 𝑺, 𝑫, 𝑺)
4. (𝑺, 𝑺, 𝑫, 𝑫)
5. (𝑺, 𝑫, 𝑺, 𝑺)
6. (𝑺, 𝑫, 𝑺, 𝑫)
7. (𝑺, 𝑫, 𝑫, 𝑺)
8 (𝑺, 𝑫, 𝑫, 𝑫)
D
S
D
9. (𝑫, 𝑺, 𝑺, 𝑺)
S
D
S
D
S
D
S
D
S
D
S
D
10. (𝑫, 𝑺, 𝑺, 𝑫)
11. (𝑫, 𝑺, 𝑫, 𝑺)
12. (𝑫, 𝑺, 𝑫, 𝑫)
13. (𝑫, 𝑫, 𝑺, 𝑺)
14. (𝑫, 𝑫, 𝑺, 𝑫)
15. (𝑫, 𝑫, 𝑫, 𝑺)
16.(𝑫, 𝑫, 𝑫, 𝑫)
Tree Diagram

To check our answer let’s use FCP
2 2 2
× × = 16 possible ways
× 2
Son or Daughter
Given:
4 children
×

In how many ways can you place 9 different books on a
shelf if there is space enough for only five books?
9 8 7
× × =
15,120 possible ways
× 6
Space on the shelf is only for 5 books
Given:
9 different books
5
×

In how many ways can Aling Rosa arrange 6 potted
plants in a row?
Using Fundamental Counting Principle
= 720 possible ways
×
× × × ×

The number of permutations of a set of n objects taken
all denoted by nPn is given by
nPn = n(n-1)(n-2)…(3)(2)(1)
= n!
A) PERMUTATION OF n DIFFERENT
OBJECTS TAKEN ALL

How many different ways can 5 parents line up to get a
new set of modules for Quarter 3?
In this kind of problem, we can use permutation of n
different objects taken all.
Example 1:
Given: n = 5
Solution: nPn = 𝑛!
5P5 = 5!
= 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
= 120 ways

In how many ways can six potted plants be
arranged in a row?
Example 2:
Given: n = 6
6P6 = 6!
= 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
= 720 ways

In how many ways can 4 contestants be arranged
in a row on the stage?
Example 3:
Given: n = 4
4P4 = 4!
= 4 ⦁ 3 ⦁ 2 ⦁ 1
= 24 ways

B) PERMUTATION OF n DIFFERENT OBJECTS
TAKEN r at a time
The number of permutations of a set of n objects taken r at a
time denoted by nPr is given by
nPr =
𝐧!
𝐧−𝐫 !
when,
a. the n objects are distinct;
b. once an object is used it cannot be repeated; and
c. order is important.
The permutation of n objects taken r at a time is denoted by
nPr. In some books, it is also denoted by P(n, r), Pn,r , or Prn . In
this learning material, we will use the first notation.

The permutation of 4 objects taken 2 at a time.
Evaluate each expression:
n = 4, r = 2
nPr =
𝒏!
𝒏−𝒓 !
4P2=
𝟒!
𝟒−𝟐 !
=
𝟒!
𝟐!
=
4 ⦁ 3 ⦁ 2 ⦁ 1
2 ⦁ 1 = 4 ⦁ 3 = 12
4P2

Evaluate each expression:
n =5, r = 5
nPr =
𝒏!
𝒏−𝒓 !
5P5= 5!
= 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
= 120
5P5
n =5, r = 0
nPr =
𝒏!
𝒏−𝒓 !
5P0 = 𝟏
5P0

How many ATM pincode can be produced with 4 digits if repetition is
not allowed?
There are 10 unique digits from 0 to 9 but the code needs only 4 digits,
then this is Permutation of n different objects taken r at a time.
Example 1:
Given: n = 10, r=4
Solution: nPr =
𝒏!
𝒏−𝒓 !
10P4 =
𝟏𝟎!
𝟏𝟎−𝟒 !
=
𝟏𝟎!
𝟔!
=
10 ⦁ 9 ⦁ 8 ⦁ 7⦁ 6 ⦁ 5 ⦁4 ⦁ 3 ⦁ 2 ⦁ 1
6 ⦁ 5 ⦁4 ⦁ 3 ⦁ 2 ⦁ 1
=10 ⦁ 9 ⦁ 8 ⦁ 7 = 5, 040
There are 5, 040 pincodes.

A class is to select a president, a vice-president, a secretary, and
a treasurer from 7 class members. How many arrangements of
class officers are possible?
Example 2:
Given: n = 7, r = 4
Solution: nPr =
𝒏!
𝒏−𝒓 !
7P4 =
𝟕!
𝟕−𝟒 !
=
𝟕!
𝟑!
=
7⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
3 ⦁ 2 ⦁ 1
=7 ⦁ 6 ⦁ 5 ⦁ 4 = 840
There are 𝟖𝟒𝟎
arrangements.

In how many ways can a grocery owner display 5
brands of canned goods in 3 spaces on a shelf?
Example 3:
Given: n = 5, r = 3
Solution: nPr =
𝒏!
𝒏−𝒓 !
5P3 =
𝟓!
𝟓−𝟑 !
=
𝟓!
𝟐!
=
5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1
2 ⦁ 1
= 5 ⦁ 4 ⦁ 3 = 60
There are 6𝟎
arrangements.

1. Five persons are to be seated in a 10-seater
AUV. In how many ways can they be seated?
Activity:
2. A pianist plans to play five pieces at a
recital. In how many ways can he arrange
these pieces in the program?

1. Five persons are to be seated in a 10-seater
AUV. In how many ways can they be seated?
Activity:
Given: n = 10, r = 5
Solution: nPr =
𝒏!
𝒏−𝒓 !
10P5 =
𝟏𝟎!
𝟏𝟎−𝟓 !
=
𝟏𝟎!
𝟓!
=
10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6⦁ 5 ⦁ 4⦁ 3 ⦁ 2⦁1
5 ⦁ 4⦁ 3 ⦁ 2⦁ 1
= 10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6 = 30,240 ways
-Determine the permutation of 10 seats taken 5 at a time.

2. A pianist plans to play five pieces at a recital. In how
many ways can he arrange these pieces in the program?
Activity:
Given: n = 5, r = 5
Solution: nPr =
𝒏!
𝒏−𝒓 !
5P5 =
𝟓!
𝟓−𝟓 !
= 𝟓! = 120 ways

1. Seven hospitals are in need of nurses. If there
are three qualified applicants, how many ways
can they be assigned?
Evaluation
2. In how many different ways can a president,
vice-president, a secretary and a treasurer be
chosen from a class of 10 students.

1. Seven hospitals are in need of nurses. If there are
three qualified applicants, how many ways can they be
assigned?
Evaluation
Given: n = 7, r = 3
Solution: nPr =
𝒏!
𝒏−𝒓 !
7P3 =
𝟕!
𝟕−𝟑 !
=
𝟕!
𝟒!
=
7 ⦁ 6⦁ 5 ⦁ 4⦁ 3 ⦁ 2⦁1
4⦁ 3 ⦁ 2⦁ 1
= 𝟕 ⦁ 6 ⦁ 5 = 210 ways

2. In how many different ways can a president, vice-
president, a secretary and a treasurer be chosen from a
class of 10 students.
Activity:
Given: n = 10, r = 4
Solution: nPr =
𝒏!
𝒏−𝒓 !
10P4 =
𝟏𝟎!
𝟏𝟎−𝟒 !
=
𝟏𝟎!
𝟔!
=
10 ⦁ 9⦁ 8⦁ 7 ⦁ 6⦁ 5 ⦁ 4⦁ 3 ⦁ 2⦁1
6 ⦁ 5⦁4⦁ 3 ⦁ 2⦁ 1
= 10⦁9⦁8⦁7 = 5, 040 ways

C) DISTINGUISHABLE PERMUTATIONS
The number of permutations of n objects of which n1 are
of one kind, n2 are of a second kind, …, and nk are of
kth kind is given by
P =
𝒏!
𝒏𝟏!⦁𝒏𝟐! ⦁…⦁𝒏𝒌!
where n = n1 + n2 + …+ nk.
Distinguishable permutation happens when some of the
objects in an arrangement are alike.

Janice is arranging books in the shelf. She has 15 books that must be
arranged. In how many ways can she arrange 6 Math, 2 English, 4
Science, and 3 Filipino books?
In this kind of problem, since there are identical elements/objects, we
will use the formula of distinguishable permutation.
Example 1:
Given: n = 15, n1=6, n2=2, n3=4, n4=3
Solution:
𝑷 =
𝒏!
𝒏𝟏!⦁𝒏𝟐! ⦁…⦁𝒏𝒌!
𝑷 =
𝟏𝟓!
𝟔! 𝟐! 𝟒! 𝟑!
There are
6,306,300
ways.
𝑷 =
15 ⦁ 14 ⦁ 13 ⦁ 12 ⦁ 11 ⦁ 10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2
6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 2 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 3 ⦁ 2
𝟓
𝑷 = 5 ⦁ 14 ⦁ 13 ⦁ 11 ⦁ 10 ⦁ 9⦁ 7
𝑷 = 6,306,300

How many different 9 letter words (real or imaginary) can be
formed from the letters in the word COMMITTEE?
The presence of 2 M’s, 2 T’s and 2 E’s reduces the number of
different words.
Example 2:
Given: n = 9, n1=2, n2=2, n3=2
Solution:
𝑷 =
𝒏!
𝒏𝟏!⦁𝒏𝟐! ⦁…⦁𝒏𝒌!
𝑷 =
𝟗!
𝟐! 𝟐! 𝟐!
There are
𝟒𝟓, 𝟑𝟔𝟎
words.
𝑷 =
9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2
2 ⦁ 2 ⦁ 2
𝑷 = 9 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2
𝑷 = 45, 360

In how many distinguishable permutations does the
number 5, 008, 443, 471 have?
The presence of 2 0’s and 3 4’s reduces the number of
different numbers.
Example 3:
Given: n = 10, n1=2, n2=3
Solution:
𝑷 =
𝒏!
𝒏𝟏!⦁𝒏𝟐! ⦁…⦁𝒏𝒌!
𝑷 =
𝟏𝟎!
𝟐! 𝟑!
There are
𝟑𝟎𝟐, 𝟒𝟎𝟎
numbers.
𝑷 =
10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2
2 ⦁ 3⦁ 2
𝑷 = 10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 5 ⦁ 4 ⦁ 3
𝑷 = 302, 400

D) CIRCULAR PERMUTATIONS
The number of circular permutations of n objects is
P =
𝒏!
𝒏
or
(𝒏 − 𝟏)!
A circular permutation is a circular arrangement of elements for
which the order of the elements must be taken into account.

Find the number of circular permutations of
seven dancers.
Example 1:
Given: n = 7
There are
𝟕𝟐𝟎
circular permutations.
Solution:
𝑷 = (𝒏 − 𝟏)!
= (𝟕 − 𝟏)!
= 𝟔!
= 𝟕𝟐𝟎

In how many ways can 12 football players be arranged
in a circular huddle?
Example 2:
Given: n = 12
There are
𝟕𝟐𝟎
Solution:
𝑷 = (𝒏 − 𝟏)!
= (𝟏𝟐 − 𝟏)!
= 𝟏𝟏!
= 𝟕𝟐𝟎
= 𝟕𝟐𝟎

Find the number of ways in which 5 people, Hazel, Beverly, Grace, Regina,
and Jhian can sit around a circular table, such that Hazel and Jhian
always sit next to each other?
In this kind of problem, using the hint term “circular table” we can use
circular permutation. But as you can see in the problem, there is a
restriction wherein Hazel and Jhian wants to sit next to each other and
that is considered as a single entity. So n = 4 but Hazel and Jhian has
their own arrangements and that is 2!.
Example 3:
Given: n = 4
There are 𝟏𝟐
Solution:
𝑷 = 𝒏 − 𝟏 ! (𝟐!)
= (𝟒 − 𝟏)! (𝟐!)
= 𝟑! 𝟐!
= 𝟏𝟐

A. Solve the following problems.
1) How many permutations can be formed from the
letters of the word COVID using all letters?
2) In how many ways can a president, vice-president,
a secretary and a treasurer be chosen from a class of 8
students?
3) In how many distinguishable permutations can be
formed using the word ACTIVITY?
4) A spinner is to be divided into four equal parts. In how
many different ways can you arrange four symbols in the
spinner?

1)Find the number of distinguishable permutations of the digits of
the number 14344.
2.) A leader wants to assign 4 different tasks to his 4 members. In
how many possible ways can he do it?
3.) How many 3-digit numbers for a lotto game can be formed from
the favorite numbers 1, 2, 5, 7, 8, and 9 if no repetition is allowed?
4.) Find the number of ways in which 5 people Ana, Baby, Carrie,
Dolly, and Eve can be seated at a round table such that Ana and Baby
must always sit together.
5.) 10 toppings for pizza are placed on the tray. How many ways can
they be arranged?
A) Solve the following problems.

It is in international summits that major world decisions
happen because of the pandemic that we are facing in the
present situation. Suppose that you are the overall in-charge
of the seating arrangement in an international convention
wherein 12 country-representative are invited. They are the
prime ministers/presidents of the countries of Canada, China,
France, Germany, India, Japan, Libya, Malaysia, Philippines,
South Korea, USA, and the United Kingdom. What style of
arrangement would you prefer? Is it linear or circular? Why?
Create your seat plan for these 12 leaders based on your
knowledge of their backgrounds. Discuss why you arranged
them that way.
Home-based Activity:Engaged!

G10-MathQ3-Week-2- Definition of Permutation

More Related Content

What's hot (20)

Similar to G10-MathQ3-Week-2- Definition of Permutation (20)

Recently uploaded (20)

G10-MathQ3-Week-2- Definition of Permutation