Detailed Lesson Plan in Mathematics 10 I.
OBJECTIVES
At the end of the lesson, lesson, 75% of of the students students will will be able to: 1. illustrate permutations of the distinct objects 2. derive the formula for finding the number of permutations of taken r at a time, n ≥ r; and 3. solve real life problems involving permutation. II.
n
objects
SUBJECT MATTER
A. Topic: Permutation Permutations s (distinct (distinct object) B. Reference: Materials Learner’s Module Grade 10, Pgs. 283 -291 C. Instructional Materials: combination lock, boxes, chalk and chalkboard, charts and colored papers. III.
PROCEDURES
Teacher’s Activity
Learners’ Activity
A. PREPARATORY ACTIVITIES
Prayer Greetings Good Morning/Afternoon, Class! Kindly check your area if there are pieces of papers or wrappers, then pick it up and throw it on the trash can. Alright. You You may now take your seats. Checking of attendance
B. MOTIVATION
Okay Class, what do you call this kind of lock? (show to them the combination lock)
That is a combination lock!
Yes! This is a combination lock. Why do you think it is called a combination lock? Why not a padlock? Or just a lock?
Because it could be locked using a combination of numbers.
Very good! I could set a combination of numbers say, 132, as my password to secure my things like my baggage when I’m travelling. Do you have this at home, class? Then what do you do to secure your things? Say, your wallet, cellphone, and other important things? Alright! Alright! It is important important for us to secure our things so that we will not lose it or get it taken away from us.
No, ma’am
Students’ answers may vary.
Do you understand, class? Yes, ma’am! Now, let’s divide div ide the class into 4. Count 1, starting from there.
1, 2, 3, 4, 1, 2,…
All number 1 will stay stay here, all number 2 stay there, … Go now to your respective group quietly in the count of 5. Okay, class we will have a game entitled “UNLOCK THE BOX” I am going to distribute a combination lock for each group. This game is simple. All you need to do is to unlock the box in 2 minutes. Take note that every member of the group should be given a chance to unlock it. There are items in the box and that will be your prize. HINT: the 3 digit that will lead to the possible code are 1,5, and 8. Do you have any clarifications? None? None, ma’am. Okay. The 2 minutes will starts now! (alarmed ring, after 2 minutes) TIME’S UP! Did all 4 groups unlock the box? Yes, ma’am! Very Good! Before you arrive at the correct code, what are the arrangements that you’ve tried? Okay. So, including the 1 st possible code (1,5,8) and the exact code (5,1,8), how many possible arrangements of the 3 digits?
185 851
581 518
815
There are 6 possible arrangements of the digits 1 5 8.
That’s right. How about if there were 8 digits in a code like your facebook account password, can you tell how many possible arrangements of your password?
Yes, Ma’am. We can tell how many possible arrangements codes are there in 8-digits but it will take a lot of time to count all the possible arrangements.
EXACTLY!
C. PRESENTATION
Now class, there is a term in mathematics that refers to the possible arrangements of things. And that is PERMUTATION! – refers PERMUTATION – refers
to any one of all possible arrangements of the elements of the given set.
Please read… read …
(reading) Yes, Ma’am.
For instance, given a set of distinct objects, we can arrange them in several ways. Like what we did with the possible arrangements of the 3-digit code. The listed arrangements are what we called ________. ________. PERMUTATIONS! Again, what are the permutations permutations of the digits 1,5, and 8?
185 851
Yes, that’s correct! Now, how are we going to find the number of permutations of distinct n objects taken r at a time? *n is the number of objects and r is the number of how many objects will be taken at a time. Class, select your partner for our next activity. Do you have now your partner?
Yes, Ma’am.
Good! Now, perform the activity entitled : LET’S FIND OUT. Here is the direction and the format of the activity, which you will write on your 1 whole sheet of paper. (post the format on the board) I will give you 5 minutes to work on that. Remember that no erasures. Do you have any clarifications? Good!
None.
581 518
815 158
(After 5 minutes) Are you done?
Yes, Ma’am.
Alright! Alright! Pass your your paper to the center center aisle and pass it to the front. In a count of 5 pass it now. n Who will come to the front and write the possible and the number of arrangements on the board? For 1 piece at a time? For 2 piece at a time? For 3? And for 4? 4? Say something something about your your answers.
r 1 piece at a time: 1, 2, 3, 4
2 pieces at a time: 12, 13, 14, s dr a c )
… ,
3 pieces at a time: 123, 124, 132, 134,
r 4
e b 3 ,
m 2
u n
,
u g
1
… 4 pieces at a time:
r ,
of
Okay, class do you have the same answer?
p 4
(
e
.
12
24
24
1234, 1243, 1342, …
Yes, Ma’am.
Very Good! Those are correct. Let’s have the summary: n r P 4 1 4 4 2 12 4 3 24 4 4 24 What do you call each arrangement?
Each arrangement is called a permutation.
Yes, it is.
Okay, let’s have the 4 th column to be the Pattern. N 4 4 4 4
r 1 2 3 4
P 4 12 24 24
Pattern. 4x1=4 4 x 3 = 12 4 x 3 x 2 = 24 4 x 3 x 2 x 1 = 24
What have you noticed on the solutions?
The first factor of each solution is 4 which is the value of n.
That’s right! right ! Any other other observations? observations? Very Good!
The succeeding factors decreases by 1 each time.
Another one? one? Very Good! Alright, the number of permutation permutation of 4 cards taken 3 at a time is 24. Since the 1 st factor is 4 (n), the 2 nd is 3 (n-1) and the last factor is 2 (n-2). )(4 2) 4 3 2 = 4 (4 1)(4 = 24 Is the answer the same in the table? Okay. The permutation of 4 cards taken 3 at a time is denoted as, P(4,3). So, the permutation of n objects taken r at a time is denoted as, _____. Excellent!
The number of factors is equal to the value of r.
Yes, Ma’am.
The permutation of n objects taken r at a time is denoted as P(n,r).
Look at the factors: (, ) = ( 1) ( 2)… (, ) = [ (1 1)] [ (2 1)] )] [ (3 1)] … [ ( 1)] ( + 1)
So, what is the last factor of P(n,r) ? Now, what is the formula in finding the number of permutations of n objects taken r at a time, P(n,r),? That’s right. (, ) = ( 1)( 2) … ( + 1 ) Example: In how many ways can 5 people arrange themselves in a row for picture taking? What is the value of n? How about the value of r? Okay, who will answer on the board? Discuss your answer.
Very Good! In the given example, did she used all the number from n=5 down to 1? Alright. Alright. Another way of writing writing P(5,5)=5 P(5,5)=5 x 4 x 3 x 2 x 1 is by 5! (read as 5 factorial). So the permutation of n objects taken all at a time is denoted by _____. Yes, it is.
Let’s try this, this , 4! = 4 x 3 x 2 x 1 = 24
The last factor of P(n,r) is ( + 1). The formula in finding the number of permutations of n objects taken r at a time, is (, ) = ( 1) ( 2)… + 1 ).
The value of n is 5. The value of r is 5. Solution: n=5, r=5 (5,5) = 5(5 1)(5 2)(5 )(5 4) *since there are 5 factors 3)(5 = 5 x 4x 3 x2 x1 = 120 There 120 ways arrangement of 5 people taking pictures in a row. Yes, Ma’am.
The permutations of n objects taken all at a time is denoted by P(n,n)=n!.
24
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3 628 800
With the same situation on the example, how can 5 people arrange themselves 3 at a time? What is n? r? Who will solve on the board?
Very good! That’s correct. Just take a look of the solution. 54321 = 21 5! = 2! = 60 What did you notice on the solution? Excellent! What is the difference between the numerator and the denominator? Is that the value of r? So, now base on the example, what is now the formula of finding the number of permutations of distinct n objects taken r at a time?
40,320 3,628,800
The value of n is 5 and the value of r is 3. Solution: = 5, 5, = 3 (5,3) = 54 543 3 = 60 60 ways that the 5 people can arrange themselves 3 at a time.
The numerator is 5! And the denominator is 2!. 5-2=3, the difference between the numerator and the denominator is 3. Yes, Ma’am. The formula of finding the number of permutations of distinct n object ! taken r at a time is (, ) = . (−)!
Excellent! That’s correct! There is a restriction that ≥ . The denominator could be 0 because 0!=1, but not negative because we can’t arrange objects NEGATIVE times. Did you get it?
Yes, Ma’am.
Alright! D. GENERALIZATION
1. How are you going to calculate the different permutations of distinct objects?
We can calculate the different permutations by using the formula ! (, ) = ; ≥ and (− )!
Excellent!
(, ) = ! and by applying the Fundamental Counting Principle and Multiply.
2. What is a permutation? That’s right! right !
Permutation refers to any one of all the possible arrangements of the
elements in the given set.
E. APPLICATION
Now, try to answer the following problems. After 5 minutes. I will call some of you to present your answer on the board. Problems: 1. Suppose you secured your laptop using a password. Later, you realized that you forgot the 5-digit code. You only remembered that the code contains the digits 1, 4, 3, j and B. How many possible codes are there? 2. Suppose that in a certain association, there are 12 elected members of the Board of Directors. In how many ways can a president, a vice president, a secretary, and a treasurer be selected from the board? 3. A dress shop owner has 8 new dresses that she wants to display in the window. If the window has 5 mannequins, in how many ways can she dress them up?
IV.
EVALUATION
Direction: Choose the number that you think is the correct answer. 1. What is the term that refers to the possible arrangement of things? 1. Combination 3. Presentation 2. Organization 4. Permutation 2. Which of the following formula is appropriate in getting the permutations of an object taken r at a time? 1. (, ) = (
!
−) − )! !
2. (, ) = (
− )!
3. (, ) = ! 4. (, ) = !
3. Which of the following equations will give the permutation of the 6-letter word PLANTS taking 4 at a time? 1. (6,4) 2.
!
=(
−)! !
(6,4) = (
−)!
3. (4,6) =
! (−)! !
4. (4,6) = (
−)!
4. Antonio received 7 paintings from his boss as a reward for his good work. He wants to hang it on his wall horizontally. In how many possible ways can he arrange it all? (show your solution on your paper) 1. 7 2. 49 3. 210 4. 5040 5. Lila bought a book from each of the 6 different genres. When she got home, she found out that there is only space for 5 books. In how many possible ways can she arrange them to fit all of the 6 books? (show your solution on your paper) 1. 35 2. 120 3. 2520 4. 5040
V.
ASSIGNMENT Think about of a situation that involves arrangements. What y ou’re going to do is to: 1. Write all the possible permutations of n object taken all at a time. 2. Write all the possible permutations of n object taken r at a time. Note: The value of your n must be greater than 4 and the value of your r must be greater than 3. Write it on your journal. For example, if you are watching athletic games. Say ten runners join a race. In how many possible ways can they be arranged as first, second, and third placers? Then answer.
JESSICA G. BLANDO
BSED – BSED – 4A, 4A, MATH MAJOR
