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Mathematics

Quarter 3 – Module 27:

Permutation of Identical Objects

and Circular Permutation

Mathematics – Grade 10 Alternative Delivery Mode Quarter 3 – Module 27: Permutation of Identical Objects and Circular Permutation First Edition, 2020

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Development Team of the Module Author: Ricky P. Landocan Editors: Aiza R. Bitanga, Bryan A. Hidalgo Reviewers: Aiza R. Bitanga, Bryan A. Hidalgo Layout Artist: Peter A. Alavanza Management Team: May B. Eclar Marie Carolyn B. Verano Marciana M. Aydinan Carmel F. Meris Ethielyn Taqued Edgar H. Madlaing Francisco C. Copsiyan

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by- step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self- check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

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What I Need To Know

This module was designed and written with you in mind. It is here to help you solve problems involving permutation of identical objects and circular permutation. The scope of this module permits it to be used in many different learning situations. The lessons are arranged to follow the standard sequence of the course but the pacing in which you read and comprehend the contents and answer the exercises in this module will depend on your ability. After going through this module, you are expected to be able to demonstrate understanding of key concepts of permutation. Specifically, you should be able to:

1. illustrate the permutation of identical objects and circular permutation; and

2. solve problems involving permutation of identical objects and circular permutation.

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___ 7. Find the number of distinguishable permutations of the letters of the word “PANAGBENGA”. A. 3,628,800 B. 518,400 C. 302,400 D. 151,

___ 8. In how many ways can three identical Mathematics magazines, two identical Statistics magazines, and four identical Science magazines be put in a stack? A. 362,880 B. 40,320 C. 1,728 D. 1,

___ 9. In how many ways can nine people be seated around a circular table? A. 362,880 B. 40,320 C. 5,040 D. 720

___ 10. How many ways can 12 people be seated around a circular table if two insist to sit next to each other? A. 7,257, 600 B. 3,628,800 C. 725,760 D. 362,

___ 11. Five different keys are to be placed on a circular key chain. How many different arrangements are there? A. 4 B. 5 C. 12 D. 24

___ 12. Seven different beads will be put together to form a bracelet with a lock. How many ways can the beads be arranged? A. 5,040 B. 2,520 C. 720 D. 360

___ 13. How many ways can eight unique beads be arranged on a chain without a clasp? A. 40,320 B. 20,160 C. 5,040 D. 2,

___ 14. In how many ways can six people be seated in a round table? A. 24 B. 120 C. 720 D. 5,

___ 15. Ten people are to be seated around a table. One of them is to be seated close to the door. How many arrangements are possible? A. 479,001,600 B. 39,916,800 C. 3,628,800 D. 362,

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Lesson

1

Permutation of Identical

Objects and Circular

Permutation

What’s In

In the previous module, you have learned that a permutation is an ordering or an arrangement of certain objects. Oftentimes, permutation problems will ask you in how many ways can you pick and arrange a certain number of objects from a larger set of objects. Let’s review our previous lesson on permutation of distinguishable objects by studying the following examples:

Example 1 In how many ways can you arrange five (5) people to be seated in a row?

Solution The diagram illustrates the five seats. Each person can be arranged in different ways.

Seat 1 Seat 2 Seat 3 Seat 4 Seat 5

In how many ways can the person be seated?

Why?

1 st person

5 ways The first person can be seated in any one of the five slots. Each slot is different from the others.

2 nd person

4 ways Since the first person is seated, the second person may occupy any of the four vacant seats.

3 rd person

3 ways The third person may occupy any of the three vacant seats.

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N O N E

N O N E

If you interchange the two letter “Ns”, you form the same word. The seemingly two different arrangements are actually one and the same arrangement. Therefore, these are considered one arrangement only.

Let’s solve the problem How many ways can the letters in the word “NONE” be arranged?

Step 1 4 P 4  4! Since there are four letters, get the permutation of the four letters.

Note: When you get the permutation of the four letters, you will be counting some words twice.

Step 2 2!

4! The permutation of the four letters will be divided by the number of ways to arrange the repeated letter.

Note: The number of ways to arrange the two letter “Ns” is (^) 2 P 2  2!.

Step 3 12 2

4! (^)   Simplify.

Therefore, there are 12 ways to arrange the letters of the word “NONE”.

To verify the answer, let us list down all the possible arrangements of the letters of the word “NONE” and remove the repeated words.

NONE ONNE NONE ENNO NOEN ONEN NOEN ENON NNOE ONEN NNOE EONN NNEO ONNE NNEO EONN NENO OENN NENO ENON NEON OENN NEON ENNO

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Example 3

How many distinguishable arrangements can be formed from the letters of the word “PAGPAPAKATAO”?

Solution Step 1

12 P 12 ^12!^ Since there are^ twelve^ letters, get the permutation of the twelve letters. Note: When you get the permutation of the twelve letters you will be counting some words twice. Step

The permutation of the twelve letters will be divided by the product of the number of ways to arrange the repeated letters. Note: The number of ways to arrange the three letter “Ps” is 3 P 3 ^3!^ and the five letter “As” is (^) 5 P 5  5!.

Step 3

Simplify.

Therefore, there are 665,280 arrangements formed from the letters of the word “PAGPAPAKATAO”.

Activity 1. Find the number of distinguishable permutations of the letters in each of the given words.

1. BAGUIO
2. REFERENCE
3. MATHEMATICS
4. BOOKKEEPER
5. STATISTICS

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Example 5 How many different eight-digit numbers can be written using the digits 1, 2, 3, 4, 4, 5, 5, and 5?

Solution Step 1 (^8) P 8  8 ! 40 , 320 Since there are eight (8) digits, get the permutation of the eight digits.

Step 2

The permutation of the eight digits will be divided by the product of the number of ways to arrange the repeated digits. Note: The number of ways to arrange the digit 4 is 2 P 2 ^2!^ and^ the^ digit^5 is 3 P 3 ^3!^. Step 3 3 , 360 12

40 , 320 2! 3!

(^8 8)   

P Simplify.

Therefore, there are 3,360 different eight-digit numbers.

Activity 2. Find the number of permutations in each situation.

1. Lisa has three vases of the same kind and two candle stands of the same kind. In how many ways can she arrange these items in a line?
2. Find the number of distinguishable permutations of the digits of the number 348,838.
3. What is the number of possible arrangements of nine books on a shelf where four Algebra books are of the same kind, three Geometry books are of the same kind, and two Statistics books are of the same kind?
4. A clothing store has shirts in four sizes: small, medium, large, and extra-large. If it has two small, three medium, six large, and two extra-large shirts in stock, in how many ways can these shirts be sold if each is sold one after the other?
5. How many different nine-digit numbers can be written using the following digits: 2,2,2,7,7,8,8,8, and 9?

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The arrangements discussed previously are linear in nature. There are some arrangements which are circular in nature such as sitting in a roundtable, making a necklace with different colored beads, and the like. The number of ways of counting associated with the circular arrangement gives rise to circular permutation ( P ).

Suppose there are five chairs around a table to be occupied by five persons A, B, C, D, and E, in how many ways can they arrange themselves?

These five persons A, B, C, D, and E can arrange themselves in 5! ways if

they are to be arranged in a row. There is a start and there is an end.

A B C D E

Seat 1 Seat 2 Seat 3 Seat 4 Seat 5 In a circular permutation, there is nothing like a start or an end. The picture illustrates the five chairs around a table. The five persons can be arranged in different ways.

https://www.houzz.com/products/6-piece-outdoor-teak- dining-set-60-round-table-5-celo-stacking-arm-chairs-prvw-vr~

How many person/s will occupy a seat?

Why?

1 st seat

1 person A particular chair is to be designated as the first seat which serves as the starting point, and to be occupied by the designated first person, say D.

2 nd seat

4 persons Since the designated first seat is already occupied by the designated first person (D), then the next chair, either clockwise or counterclockwise, can now be occupied by any of the four remaining persons A, B, C, or E. Let us say B occupies the second seat.

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Example 6 Ten boy scouts are to be seated around a campfire. How many ways can they be arranged?

Solution

P  

Therefore, there are 362,880 ways to arrange the ten boy scouts around a campfire.

Example 7 Eight people are to be seated at a roundtable. One of them is to be seated close to the window. How many arrangements are possible?

Solution P  n!  8!  40 , 320 Note: If n objects on a circle are arranged relative to

a fixed point, then there are n! permutations even

though the objects are on a circle, the permutations are linear since a reference point has been established. Therefore, there are 40,320 arrangements possible.

Example 8 How many different ways can four keys, no two of which are the same, be arranged on a key-ring that has a clasp?

Solution 12 2

P  n!^   

Note: This is no longer a case of an ordinary circular permutation since objects are arranged with respect to a fixed point, the clasp, and is reflective. In this case, there are 2

n! permutations.

Therefore, there are 12 ways to arrange the four keys, no two of which are the same, on a key-ring.

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What’s More

Now, it’s your turn to answer the following activities. Activity 4. Matching Type: Match each problem at the left to the permutation notation at the right to answer the said problems. Write the letter of the correct answer on the blank before the item number. PROBLEM PERMUTATION NOTATION ___ 1. How many ways can six beads, no two of which are the same, be arranged on a bracelet without clasp? ___ 2. Using the digits 5, 6, 6, 6, 6, 6, 7, 7, 8, and 9, how many 10-digit numbers can be formed? ___ 3. In how many ways can six people be seated in a round table? ___ 4. In how many ways can five identical red cards, two identical blue cards, and three identical black cards be arranged in a row? ___ 5. How many ways can six keys, no two of which are the same, be arranged on a key ring with a clasp?

A.

B. ( 6  1 )!

C.

D.

E.

F.

Activity 3. Find the number of permutations of the given objects in each of the following situations.

1. How many seating arrangements are possible for five people at a roundtable?
2. In how many different ways can four keys, no two of which are the same, be arranged on a key-ring that has no clasp?
3. Twelve beads, no two of which are the same, are to be strung in a necklace with a clasp. In how many ways can it be done?
4. How many ways can five boys and five girls be seated alternately at a circular table?
5. How many ways can seven people be seated at a roundtable relative to the door of a room?

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What I Have Learned

Summing up, let us list down what you have learned in our discussions.

The number of distinguishable permutations of 𝑛 objects of which n 1 are alike and one of a kind,^ n 2 are alike and one of a kind, …,^ nk are

alike and one of a kind, the number of distinguishable permutations is:

n 1 n 2 nk

n .

If n objects are arranged in a circle, then there are n

n!

or  n  1 !

permutations of the n objects around the circle.

If n objects on a circle are arranged relative to a fixed

point, then there are n! permutations. Even though the

objects are on a circle, the permutations are linear since a reference point has been established.

If n objects on a circle are reflective, then there are

( n  1 )! permutation.

What I Can Do

In this part of the module, you will apply the concepts of solving permutation with different restrictions. Consider the situation:

There are three identical red books, one white book, and one black book on the shelf. How many ways can the books be arranged if the white and black books are separated?

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STEP 1 Find the total number of arrangements. 3!

Ptotal 

Ptotal  20 ways

STEP 2 Find the total number of arrangements if the white and black books are NOT separated. Note: If the white and black books are not separated or together, then they are considered as one and can be arranged in 2 P 2 ^2 !^2^ ways.

Pnot separated  

Pnot separated  8 ways

STEP 3 Find the difference between the total number of arrangements and the number of arrangements if the white and black books are not separated or together.

Pseparated  Ptotal  Pnot separated

Pseparated  20  8 Pseparated  12 ways

Therefore, there are 12 ways to arrange the books if the white and black books are separated.

Congratulations, I know that you are ready to apply what you have learned in this module.

Activity 6.

1. In how many ways can five people be arranged in a circle such that two of them sit together?

STEP 1 Find the total number of arrangements. Note: Two people who sit together are considered as one. STEP 2 Find the number of arrangements of the two people who sit together. STEP 3 Find the product of the total number of arrangements and the number of arrangements if the two people sit together.