GRADE 7
LESSON
PLAN
School:
San Pedro National High School
Grade Level:
10
Teacher:
Ms. Lorraine A. Retuya
Learning Area:
Mathematics
Teaching Dates
and Duration:
Applicant Code:
April 24, 2024/ 40 minutes
Quarter:
4TH QUARTER
JHS1Pulilan_306709_Math001
A DETAILED LESSON PLAN IN MATHEMATICS 10
I.
LEARNING OBJECTIVES
Content Standards
Performance Standards
Learning Competencies/Code
Objectives
II.
The learner demonstrates understanding of key concepts of combinatorics
and probability.
The learner is able to use precise counting technique and probability in
formulating conclusions and making decisions.
The learner is able to illustrates the permutation of objects. M10SP-IIIa-1
At the end of the lesson the students should be able to:
 Illustrate the permutation of objects;
 Solve problems involving permutations.
SUBJECT MATTER
TOPIC
REFERENCES
MATERIALS
III.
SUBJECT MATTER

PRELIMINARY ACTIVITIES
PERMUTATION
Realistic Math: Basic, Beyond, Breakthroughs 10
pp. 162-175
Calculator and PowerPoint Presentation
TEACHER’S ACTIVITY

Prayer
Before we start our discussion, I want our class
president to lead the prayer.
STUDENTS’ ACTIVITY
Let’s vow our heads and feel the presence of the
Lord,
In the name of the Father, of the Son and of the
Holy Spirit. Ament.
Lord, bless each and every student here, as well as
our teacher, as we embark on another day of
learning. Please give us clear minds and open
hearts so we can grasp the knowledge and lessons
set before us. Help us to be attentive, and let Your
wisdom guide us in our studies. Amen

Greetings
Good Morning Class!
So before you take your seats, please pick the pieces
of trashes around you.
Good Morning Ms. Retuya, it’s nice to see you.
Thank you Ma’am.
You may now take your seats.

Attendance
Let’s check the Attendance. May I know who’s
absent today?
That’s Great!
None Ma’am.

Review
So, may I ask anyone from the group who can tell
me your previous discussion in mathematics?
Alright, from the previous topic we say that there are
two equations of circles, what are those?
Very good, who can tell me the equation of a circle
(0,0) or at the origin? How about equation of circle
at (h,k)?
Fantastic do you have any clarification regarding the
previous discussion?

The equation of a circle when the center is at the
origin and the other is when the center is at (h,k)
ma’am.
The standard equation of a circle at (0,0) is
𝑥2 + 𝑦2 = 𝑟2
While the standard equation of a circle at (h,k) is
(𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2
None ma’am.
Motivation
So, before we proceed to our discussion I have here a
multiple dial padlock that requires a three-digit
passcode in order to unlock. Can you guess with me,
what is the passcode of this padlock given the
following clues?



Graphs and Equation of Circles ma’am.
Clues:
There is no Roman Numeral for this number.
It is the only prime number preceding a cube.
A paper cannot be folded more than this number of
times. Try it!
We know that the passcodes for this lock includes the digit
0,7 and 9. But how many different three-digits passcodes can
be formed using the digits 0,7 and 9?
Yes, ma’am.



0
7
9
079
097
709
790
907
970
Let us enumerate all the passcodes using 0, 7 and 9?
As what you have mentioned, there are 6 possible
arrangements of these three numbers.
Let’s try each of these passcodes. Which arrangement do you
think will open this lock?
Ma’am, the padlock was unlocked by 907.
Very good! The passcode for this padlock is 907.
 Discussion
In our day to day experiences there are a lot of things that
may be arrange into ways. Like the numbers 0,7 and 9. But
the question is how we will know how many arrangements
we can be able to form in such cases.
In probability we call those arrangements as
PERMUTATION.
May you please read the meaning of Permutation on the
screen?
Thank you! As we say, Permutation is knowing how many
arrangements are possible for an object if we are to order
them.
Permutation – is the set of all possible
arrangements of an object where order is
important.
There are 3 different cases to know the permutation of an
object:
FIRST CASE: The permutation of n objects taken n at a
time
𝑛!
SECOND CASE: The permutation of n objects taken n at a
time but with identical objects.
𝑛!
𝑘1 ! 𝑘2 ! 𝑘3! …
THIRD CASE: The permutation of n objects taken r at a
time.
𝑛!
𝑛𝑃𝑟 =
(𝑛 − 𝑟)!
In these cases, Factorial Notation is used to solve the
Permutation.
Factorial Notation:
1! =1 =
2! =2 ∙ 1 =
3! = 3 ∙ 2 ∙ 1 =
4! = 4 ∙ 3 ∙ 2 ∙ 1 =
5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =
6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =
7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =
8! = 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =
9! = 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =
10! = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 =
1
2
6
24
120
720
5,040
40,320
362,880
3,628,800
As a special case, we define 0! = 1
Example 1.
In how many ways can you arrange the letters of the
following word (real or not):
a. PEACE
b. LOVE
c. EXPLORE
Do you have any questions?
Ok, so what if I ask you to arrange the letters of my name
LORRAINE how many possible arrangements do you think
are there?
Identical letters are already found in the name LORRAINE,
but in those cases there is a certain formula for that.
SECOND CASE: The permutation of n objects taken n at a
time but with identical objects.
𝑛!
𝑘1 ! 𝑘2 ! 𝑘3! …
Example 2. How many permutations are there in the letters of
the name LORRAINE?
R= repeated twice
a. 5! = 120
b. 4! = 24
C. 7! = 5,040
8! 40, 320
=
= 20, 160
2!
2
Thus, we say that there are 20, 160 permutations in the
letters of the name LORRAINE.
Now your turn:
Example 3. Determine the number of distinct permutations
that can be formed (real or not) using all the letters of the
following names.
11!
39,916,800
= 34, 650
24∙24∙2
9!
3,62,880
b. 2!2!2! = 8 = 45,360
a. 4!4!2! =
a. Mississippi
b. Committee
Very Good!
Example 5.
In how many ways can 10 students be arranged in a row?
10P10
= 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2x 1 = 3, 628, 800
∴ There are 3, 628, 800 ways can 10 students be
arranged in a row.
Very Good! So, in this classroom if these 10 students are to
be arranged, you’ll get 3, 628, 800 possible ways. Can you
imagine that? I guess you’ll get dizzy finishing that 3 Million
plus arrangements.
However, in some cases. Not all objects are taken at the same
time. What if in a class there are only 8 chairs available but
there are 10 students that are to be seated. How many
arrangements are possible?
SECOND CASE: The permutation of n objects taken r at a
time.
𝑛!
𝑛𝑃𝑟 =
(𝑛 − 𝑟)!
Example 4: How many ways can you arrange 10 students in a
row of 8 armchairs?
What do you think which of the formulas are we going to
use?
𝑛!
𝐧!
𝑛𝑃𝑛 = 𝑛! or
or 𝐧𝐏𝐫 = (𝐧−𝐫)!
𝑘 !𝑘 !𝑘 …
1
2
𝑛𝑃𝑟 =
𝑛!
(𝑛 − 𝑟)!
3!
Very Good! So, in here, we are to apply the formula.
n=10 (since there are 10 students to be arranged)
r=8 (since 8 armchairs are only available)
10!
10! 3,628,800
=
=
(10 − 8)! 2!
2
= 1, 814, 400
∴ There are 1,814,400 ways can 10 students be
arranged in a row with only 8 chairs available.
10𝑃8 =
None ma’am!
Yes Ma’am!
Do you have a question?
Is everything clear?
Example 5: From the digits 0-9, how many three-digit whole
numbers of different digits can be formed?
Which formula are we going to use?
𝑛!
𝐧!
𝑛𝑃𝑛 = 𝑛! or
or 𝐧𝐏𝐫 = (𝐧−𝐫)!
𝑘1 !𝑘2 !𝑘3! …
𝐧𝐏𝐫 =
𝟏𝟎𝐏𝟑 =
𝐧!
(𝐧 − 𝐫)!
𝟏𝟎! 𝟏𝟎 ∙ 𝟗 ∙ 𝟖 ∙ 𝟕 ∙ 𝟔 ∙ 𝟓 ∙ 𝟒 ∙ 𝟑 ∙ 𝟐 ∙ 𝟏
=
𝟕!
𝟕∙𝟔∙𝟓∙𝟒∙𝟑∙𝟐∙𝟏
𝟏𝟎 ∙ 𝟗 ∙ 𝟖 = 𝟕𝟐𝟎
Very good, so would you mind applying the given in our
formula?
∴ There are 720 3-digit numbers can be formed
from the digits 0-9.
Well done! Fantastic!
𝐧𝐏𝐫 =
This time, your turn!
Example 6: The top 3 of the 12 regional champions
competing in the final round of a math competition will
receive gold, silver, and bronze medals. How many possible
winning orders are there in the competition?
𝐧!
(𝐧 − 𝐫)!
𝟏𝟐!
𝟏𝟐!
=
(𝟏𝟐 − 𝟑)!
𝟗!
𝟏𝟐 ∙ 𝟏𝟏 ∙ 𝟏𝟎 ∙ 𝟗 ∙ 𝟖 ∙ 𝟕 ∙ 𝟔 ∙ 𝟓 ∙ 𝟒 ∙ 𝟑 ∙ 𝟐 ∙ 𝟏
=
𝟗∙𝟖∙𝟕∙𝟔∙𝟓∙𝟒∙𝟑∙𝟐∙𝟏
𝟏𝟐 ∙ 𝟏𝟏 ∙ 𝟏𝟎 = 𝟏, 𝟑𝟐𝟎
∴ There are 1,320 possible winning orders in the
competition.
𝟏𝟐𝐏𝟑 =
 Application
This time let me ask you to do this, you may try answering it
on your notebook. But I need someone who will show his/her
answer on the board.
a. How many distinct 9-digit numbers can be
formed using the digits 1,2,1,2,3,1,2,3,4?
b. How many four-letter words (real or not) are
possible from the letters TUESDAY?
c. A minibus at the terminal can load up to 24
passengers. In how many ways can 8 passengers
waiting at the terminal could sit on the bus?
𝟗!
𝟏𝟐!
a. 𝟑!𝟑!𝟐! = 𝟗!
b. 7! = 5,040
𝟐𝟒!
𝟐𝟒!
𝟐𝟒∙𝟐𝟑∙𝟐𝟐∙𝟐𝟏∙𝟐𝟎∙𝟏𝟗∙𝟏𝟖∙𝟏𝟕∙𝟏𝟔!
c. 𝟐𝟒𝑷𝟖 = (24−8)! = 16! =
16!
= 𝟐𝟗, 𝟔𝟓𝟒, 𝟏𝟗𝟎, 𝟕𝟐𝟎
Very Good!
Do you have question?
 Generalization
So, to sum it up, May I ask you what is Permutation?
Permutation – is the set of all possible
arrangements of an object where order is
important.
Very Good! There are 3 cases in getting permutation of objects FIRST CASE: The permutation of n objects taken
can you give me one and state their formula?
n at a time
𝑛!
SECOND CASE: The permutation of n objects
taken n at a time but with identical objects.
𝑛!
𝑘1 ! 𝑘2 ! 𝑘3! …
Very Well said!
THIRD CASE: The permutation of n objects
taken r at a time.
𝑛!
𝑛𝑃𝑟 =
(𝑛 − 𝑟)!
Values Integration:
In your own words, what do you think is the importance of
In life, there are things that should be done
having the sense of order in life?
following a certain order to have better results.
IV.
EVALUATION
Do as indicated:
A. Determine the number of distinct permutations that can be formed using all the letters of
the following words:
1. BALLESTEROS
2. SANGKATAUHAN
3. MANGANTERAN
B. Solve the following:
1. How many different 4-letter arrangements are there from the letters PINOYKA?
2. At a bus stop, a minibus with 6 vacant seats accepted 4 additional passengers to
enter the bus. In how many ways can they be seated?
3. In how many ways can 8 children sit in a row of 8 chairs in their classroom?
V.
ASSIGNMENT
A. Make a list of the names of 10 important people in your life and find the permutation of
all the letters in their names. Show your solution in each.
B. Give the permutation of these names listed in A given that only 3 letters will be taken at
a time.