The document is a detailed lesson plan for a mathematics class on permutation. It outlines the objectives, content, materials, and procedures for the lesson. The lesson will teach students about permutation rules including n!, nPr, and arrangements of distinct objects. Example problems are provided to demonstrate each rule, and students will complete activities in groups to practice the rules and verify their understanding.

1. A Detailed Lesson Plan
In Mathematics 10
October 12, 2015
Ruby Rose Ann B. Panganod Ms. Catalina B. Gayas
Student Teacher
I. Objectives
Given several activities,the students should be able to do the followingwith at least 80%
proficiency:
a. To define the permutation of n objects;
b. To identify the rules of permutation;
c. To use the formula for finding the permutation of n objects taken r at a time;
d. To use the formula of circular permutation;
e. To solve problems using the different rules of permutation.
II. Content and Materials
a. Topic: Permutation
b. References: Pelingon, J., Petilos, G., Gayas, C., Fundamentals of Mathematics,
pp. 115- 120.
Davison, et. al, Pre- Algebra Course 3, pp. 324-348.
c. Materials: cartolina used as cards, cartolina for visual aids, box with numbers,
III. Procedure (Deductive Method)
Teacher’s Activity
A. Review
Now, I have here a box, and each one of
you will pick a paper. Each paper has a
number on it.
Does everything have their number?
Very good. Now, I have here another
box and I will pick a number, and
whoever has the same number that I
had picked will answer the following
questions.
So for question number 1;
Students’ Activity
Yes Maam.

2. A woman has 8 skirts and 8
blouses. In how many different
attires may she appear?
So, I’ve picked number 26. Who picked
number 26? Yes Anna?
Very good Anna. What rule did you use?
That’s right Anna.
For question no.2,
A house has three doors, in how
many ways can a person enter
and leave by another?
So I’ve picked number 16. Who picked
number 16? Yes Vince?
Very good Vince. For question no.3,
Three coins are tossed at the
same time. In how many ways can
they can up?
So I’ve picked number 34. Who has
number 34? Yes Shiela?
Very good Shiela. Lets move on to
question no. 4,
A student plans to buy one of the
following: a novel, a ballpen, and a
notebook. If there are 15 choices for
the novel, 20 choices for the ballpen
and 15 choices for the notebook, how
many choices does the student have?
So I’ve picked no.3. Who got number 3?
Yes Ann?
Very Good Ann.
Again, what is addition rule? Yes
Geraldine?
The woman has 64 attires to wear.
The rule that I use is Multiplication rule.
There are 9 ways to enter and leave the
house.
Three coins can come up in 8 possible
ways.
The student has 50 choices.
Addition rule states that the number of
ways of selecting k mutually disjoint sets is

3. Very well said Geraldine. How about
Multiplication rule? Yes Marvin?
That’s right Marvin!
Its good that you still remember our
lesson about Counting techniques.
B. Motivation
Now, you will form a group with five
members. So this row will be group 1,
this row will be group 2, group 3, group
4, group 5, group 6, group 7 and group
8. Form a circle now. Faster.
Are you in your groups now?
Very good. Now, I will give you 4 cards
with the letters M, A, T and H written
on each card.
Do you have your cards now?
Now I want you to put your card on the
table face down. Shuffle itthree times.
Done?
Now, from those 4 cards, choose 3
cards and face that card up. Now, you
have three letters. In a 1 whole sheet
of paper, I want you to record the
possible arrangement of those 3
letters. I will give you five minutes to
do that. Understood?
(After 5 minutes)
Are you done?
simply by adding the number of elements
of each set.
Multiplication rule states that if there are
m ways of performing the first step and n
ways in performing the second step, then
there are m.n ways of performing the steps
in the given order.
Yes maam.
Yes maam.
Yes maam.
Yes maam.
Yes maam.

4. Now, using all the cards, I want you to
record all the possible arrangements
of the four cards. I will give you 5
minutes to do that.
(After 5 minutes)
Are you done?
Now, Using all four cards, how many
arrangements do you have?Yes group
1?
How about the other groups, do you
have the same answer with group 1?
Okay. Later we will check if your
answers are correct. How about the
three letters, how many possible
arrangements do you have?Yes group
5?
Are your answers the same with their
answer?
Okay. We will check later if your
answers are correct.
C. Statement of the Problem
Now class, say for example, you were
riding on a bus with 2 of your friends
and there were 3 vacant seats ina row.
In how many ways can you arranged
yourself?
So I need three volunteers here in
front.
Yes Kathryn, Jane and Roel.
Say for example, Kathryn, Jane and
Roel were on the bus, and these are
the vacant seats.
Yes maam.
We have listed 24 possible arrangements
of the 4 letters.
Yes maam.
We have listed 6 possible arrangements of
the 3 letters.
Yes maam.

5. So one possible arrangement is that
Kathryn is beside Jane and Roel, or we
can represent it in symbols. So one
possible arrangement is JKR. Who can
give me another arrangement?
Yes, Jollie Mae?
Very Good. In symbols, RJK. Who can
give another possible arrangement?
Yes Mealone?
That’s right. Another one? Yes
Rommel?
Very good Rommel. How about you,
Lolita?
Very Good. Is there any possible
arrangement that was not mentioned?
Yes Fatima?
Very Good. Any possible arrangement
that was not mentioned? None?
Lets list down the possible
arrangements you mentioned.
JKR JRK
RJK RKJ
KRJ KJR
Okay. Jane, Kathryn and Roel, you can
now take your seats. Now, how many
possible arrangements are there?
Yes Anna?
Very good. So there are six possible
arrangements for three people sitting
on 3 seats on a bus.
How about if there were 8 people in a
bus? Do we have to list all the possible
Another possible arrangement is Roel,Jane
and Kathryn.
Another possible arrangement is Kathryn,
Roel and Jane.
Jane, Roel and Kathyrn is a possible
arrangement.
Roel, Kathryn and Jane is also a possible
arrangement.
Another possible arrangement is Kathryn,
Jane and Roel.
There are six possible arrangements.

6. arrangements? Do youthink itwill take
us a lot of time?
D. Statement of the
Generalization
Now class, instead of listing all the
possible arrangement of an object,
mathematics has an easy way of solving
problems, which is concerned with
arrangements.
Okay class, kindly read;
“Permutation refers to any one of all
possible arrangements of the elements
of the given set.”
For instance, given a set of distinct
objects, we can arrange them in one of
several ways. Like what we did with the
possible sitting arrangements of Roel,
Jane and Kathryn. The listed
arrangement are the permutations of
the distinct objects.
Now, lets discuss the rules of
permutation.
Kindly read, Marie?
Thank you Marie.
Class, n! = n(n-1)(n-2)(n-3) … 3.2.1. Say
for example, 5.4.3.2.1=5! and we read
this as “Five factorial”. So 5! = 120.
In our example a while ago, how many
distinct objects do we have? Yes
Joevhan?
And what are they Joevhan?
Very good Joevhan. Remember class
that the object that we are talking
about is the subject that is being
“Permutation refers to any one of all
possible arrangements of the elements
of the given set.”
“Rule no. 1: The number of permutations
of n distinct objects arranged at the same
time is given by n!”
We have 3 distinct objects.
They are Jane, Roel and Kathryn.

7. permuted. It may be an animal, a
person, a letter, or any other things.
Going back to our example, we have
three objects, so to find the possible
permutations, we will have 3!.
3! = 3.2.2
= 6
3! Is equal to six. Is it the same to our
answer a while ago?
So instead of listing all the possible
permutations of an object, we can use
n! in order to find on how many ways
can we arranged n objects.
Understood?
Lets proceed to rule no. 2. Yes, Joshua
kindly read?
Thank you Joshua.
Say for example, we have 5 passengers
and there were only 3 vacant seats. In
how many ways can we arranged the 5
passengers? Rule no.2 can answer this
question.This means that we will take 5
passengers 3 at a time or 5P3.
Substituting to the formula, we have,
5P3 =
5!
(5−3)!
=
120
2
=60
So there are 60 possible arrangements
of taking 5 passengers 3 at a time.
Understood?
Next is Rule No. 3. Yes Reslee?
Yes maam.
Yes Maam.
“Rule no.2: The number of possible objects
taken r at a time is given by
nPr =
𝑛!
( 𝑛−𝑟)!
Yes Maam.
“Rule no. 3: the number of ways of
arranging n objects of which n1 are of the

8. Thank you Reslee.
Say for example, in how many ways can
3 blue bulbs, 5 red bulbs and 2 green
bulbs be arranged in a row?
This is the same as arranging 10 objects
of which3 are alike,5 are alike and 2 are
alike. Hence, the number of possible
arrangements is
10!
3!5!2!
= 2 520
Understood?
Lets proceed to rule no. 4. Kindly read,
Oscar?
Thank you Oscar.
Rule no. 4 is also called circular
permutation. In circular permutation,
we are dealing with circular
arrangements. Say for example, we
want to find out in how many ways can
we arrange 8 trees around a circular
garden, which is given by (8-1)! = 7! =
7.6.5.4.3.2.1 = 5 040.
Understood class?
E. Inference
Using the rules of permutation, lets
answer the following questions.
same kind, n2 are of the same kind, …, nk
are of the same kind is given by
( 𝑛
𝑛1 ,𝑛2,…,𝑛 𝑘
) =
𝑛!
𝑛1 ! 𝑛2!.….𝑛 𝑘!
Yes maam.
Rule no.4: The number of arrangements
of n distinct objects around a circle is
given by (n-1)!.
Yes Maam.

9. 1. In how many ways may the letters
of EMAIL be arranged? How many
of these starts with letter A?
Now, what rule are we going to use in
order to solve thisproblem? Yes Maine?
Very good Maine. So now, how many
distinct objects are in the word EMAIL?
Yes Alden?
That’s right Alden. Now, according to
rule no. 1, EMAIL can be arranged in 5!
= 5.4.3.2.1 = 120 ways.
Now, among these 120 ways, how
many of these starts with letter A?
We can solve this by assigning letter A
as the first letter,and A is already fixed
on that position. So now,
A __ __ __ __
Now, how many lettersdo we have now
aside from A? Yes Nikka?
Very good. So now, we are dealing with
4 letters. Its like dealing with 4 distinct
objects. So using rule number 1, how
many arrangements starts with letter
A? Yes Joshua?
Very good Joshua.
Lets proceed to question no.2.
2. How many 3 digitnumbers can be
written from the digits 1,2,3
and4?
What rule can we use to answer the
second question? Yes Roel?
Why?
The rule that fits to the problem is rule
no.1.
There are 5 distinct objects.
We have 4 letters left.Letter E, M, I, and
L.
4! = 4.3.2.1 = 24. There are 24
arrangements that start with letter A.
We can use rule no.2 to answer the
question.
Because its like taking 4 numbers 3 at a
time.

10. Very good Roel. Who wants to answer
the question on the board? Yes
Cristina?
Very good Cristina.
3. In how many ways can the letters
of the word CONDENSED be
arranged in a row?
What rule can we use to answer this
question? Yes Jesie?
Okay. The word CONDENSED has 9
letters. C, O and S appeared once while
N, D and E appeared twice. Hence, the
number of permutation is given by,
( 9
1,1,1,2,2,2
) =
9!
1!1!1! 2!2!2!
Who wants to answer this on the
board? Yes Daisy?
That’s right Daisy. Lets move –on to
question no.4.
4. In how many ways can we
arranged three personsin a round
table?
Who wants to answer on the board?
Yes Jesie?
Very Good Jesie.
4P3 =
4!
(4−3)!
=
24
1
= 24
There are 24 three digitnumbers that can
be formed using the numbers 1,2,3 and
4.
We can use rule no.3 to answer the
question.
9!
1!1!1! 2!2!2!
=
362880
8
= 45360
There are 4560 ways in arranging the
letters of the word condensed.
Using rule no.4, (3-1)! = 2! = 2. There are
two ways to arrange 3 persons in a round
table.

11. A
AB
B
C C
A
B
Fig.1 Fig.2
BC
A
AB
F. Verification
Now, lets take a look at this. Say for
example, the three persons are A, B and
C.
What can you observed? Yes Joshua?
Verygood Joshua. So it means that figure
1 and figure 2 are not really identical.
Thus moving 3 persons at the same time
around the table does not result in a
different permutation. Now, what
happens if from figure 1, we change the
position of B and C and just fix the
position of person A?
This arrangement is already different to
from the arrangement in fig. 1. Now,
from fig.1, what if we fix the position of
person B and interchange the position of
A and C ?
The two arrangements are not really
distinct.
Figure 3
Figure 4
C

12. A A
B BC C
Now, what can you observed? YesNikka?
Very good Nikka. Figure 3 and Fig. 4 does
not yield into different permutation.
Now, what if at this time, from Figure 1,
we interchange the position of A and B
and fix person C?
Now, what can you observe? Yes
Marvin?
Very good Marvin. If we rotate fig.3, it
will look like figure 4 and figure 5. Thus,
there are only 2 possible permutation in
arranging 3 persons at a round table.
Now, I will group you into 4. So this will
be group 1, group 2, group 3 and group
4. In each group, I will give you a manila
paper and a pentel pen. You will write
your answers on the manila paper.
Now, lets check if your answer to our
activity a while ago is correct. If we want
to find out how many possible
arrangements can we make out of the
Fig.3 and Fig. 4 are identical.
Figure 5 is also identical with Figure 3.
CA
B
Figure 5

13. four letters, what rule are we going to
use to solve this problem? Yes Jane?
Very good Jane.
Now, using rule no.1, is your answers
correct?
So in finding how many ways can we
arranged 4 letters, it is simply 4! = 24.
And the number of ways in arranging 3
letters is 3! =6 ways.
Very good class. Lets give yourselves a
round of applause.
We will use rule no.1.
Yes maam.
IV. Evaluation
Answer the followingquestionsand state what rule is applicable to solve the problem.
1. In how many ways can the letters w, x, y, and z be arranged in a row?
2. In how many ways can you arrange the lettersof the word FACEBOOK? How many
of these starts and ends with a consonant?
3. How manywayscan 4 membersof a familybe seatedinatheatre if the motheris
seatedonthe aisle?
4. There are 720 ways forthree studentstowinfirst,second,andthirdplace ina
debatingmuch.Howmanystudentswere competing?
5. In how many ways may a number of 5 figures be written from the digits 0 to 8
inclusive without repeating any digit?
6. How many numbers between 2000 and 5000 can be formed from the digits 2, 4,
3, 9 and 7? If repetition is not allowed?
7. In how many ways can seven people be sitted on a round table?
V. Assignment
In a one whole sheet of paper, answer the following completely and neatly.
1. In how many ways can a party of 4 girls and 4 boys be placed at a round table so that
boys and girls alternate?
2. In how many ways can 8 people be seated at a round table
a. If a certain three insists on sitting next to each other,
b. If they refuse to sit next to each other?