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Mathematics 10, Quarter 3 - Module 1, Illustrating Permutation of Objects, , This instructional material was collaboratively developed and, reviewed by educators from public schools. We encourage teachers, and other education stakeholders to email their feedback,, comments, and recommendations to the Department of Education, at [email protected]., , We value your feedback and recommendations., , 1

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Lesson, , 1, , Illustrating Permutation of, Distinct Objects, , What’s In, Determining the different possibilities of an event requires counting. There are four, techniques in counting events namely making a table, tree diagram, systematic listing and, fundamental principle of counting., Making a table is a technique where values or different possibilities are tabulated., Tree diagram is another technique which uses line segments originating from an event to an, outcome. This is a picture of all possible outcomes when an event is unfolded., Systematic listing is a counting technique that involves complete list of all possible, outcomes., Fundamental counting principle is a counting technique in which if two events are, independent and one event occurs in a ways, and the other event occurs in b ways, then,, these events occur in ab ways., , Example:, There are 4 cyclists in a race. In how many ways will they be arranged as first,, second, and third placers?, To answer this problem, we can use any of the four counting techniques mentioned, above., Making a table, Number of Cyclists, 4 (A,B,C,D), , Possible Arrangements, 3 cyclists at a time (for first,, second, third):, ABC, ABD, ACB, ACD,, ADB, ADC, BAC, BAD,, BCA, BCD, BDA, BDC,, CAB, CAD, CBA, CBD,, CDA, CDB, DAB, DAC,, DBA, DBC, DCA, DCB, , Tree diagram, , 2, , Number of Arrangements, 24

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Thus, there are 24 possible arrangements., Systematic listing, There are only four cyclists: A, B, C and D. There are four cyclists to choose from for, the first place, three cyclists to choose from for the second place and two cyclists to choose, from for the third place., The actual list of all possible arrangements are ABC, ABD, ACB, ACD, ADB, ADC,, BAC, BAD, BCA, BCD, BDA, BDC, CAB, CAD, CBA, CBD, CDA, CDB, DAB, DAC, DBA,, DBC, DCA, DCB., Therefore, there are 24 possible arrangements. Fundamental, counting principle, first placer, , second placer, , third placer, , Take a second look at the problem above., 1. Describe the arrangement of cyclists as first, second, and third placers? Is the order, or arrangement important?, 2. Is the list of possible arrangements of cyclists that are made is complete and actual?, 3. What do you call each possible arrangement of cyclists?, The arrangement of cyclists is in absolute order. The arrangement is important; that, is, a different arrangement means a different result. Each possible arrangement of cyclist is, called permutation., What is Permutation?, Permutation refers to the different arrangement of objects in a definitive manner., The order of the objects is important. There are two types of permutations. These are, permutations of objects in a line and permutations of objects in a circle. Permutation of, distinct objects and permutation of non-distinct objects are the two kinds of permutations of, objects in a line. But in this lesson our focus is permutation of distinct objects., What is permutation of distinct objects?, Permutation of distinct objects refers to the different arrangements of distinct, objects in a line. It is the different arrangements when no objects are identical or the same., Examples:, 1. Arranging 4 different potted plants in a row, - What makes this a permutation of distinct objects is that, 4 different potted plants, are distinct objects and are arranged in a line., 2. Getting the possible arrangements of letters of READ, 3

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- The four letters in READ are all distinct letters and are arranged in a line., 3. Choosing president, vice president, secretary and treasurer from 12 members in a, \A president can’t be a vice president or a secretary or a treasurer at the same, time so meaning, 4 different officers are to be chosen from 12 members., , Let’s find out how to illustrate permutation of distinct objects., Situation/ Activity, Example 1:, Arranging 3 different, mathematics books, in a shelf, , Number of, Objects, 3, , Discussion of the Possible Arrangements, , 3 books at a time:, Suppose the three different mathematics books, have titles, Algebra, Geometry, and Statistics. Let, us code them with letters A, G, S respectively., The possible arrangements are:, AGS, ASG, GSA, GAS, SAG, SGA, With the use of Fundamental Counting Principle:, , Thus, there are 6 possible arrangements of, mathematics books in a shelf., Example 2:, Arranging 4 different, potted plants in a row, , 4, , 4 potted plants at a time: the potted plants can be, arranged according to height, or according to kind,, according to appearance, or any basis we want. For, instance, 4 potted plants are coded with A, B, C, D., the possible arrangements are:, ABCD, ABDC, ACDB, ACBD, ADBC, ADCB, BCDA,, BCAD, BDAC, BDCA, BACD, BADC,, CDAB, CDBA, CABD, CADB, CBDA, CBAD,, DABC, DACB, DBCA, DBAC, DCAB, DCBA, With the use of Fundamental Counting Principle:, Therefore, there are 24 possible arrangements of 4, potted plants in a row., , 4

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Example 3:, Electing a, Mathematics club, president,, vicepresident and a, secretary from 10, members, , 10, , 3 members at a time:, Suppose the names of the 10 members:, J. Gubaton, A. Perez, I. Macamay, R. Cruz, N. Torres, F. Leon, T. Fernandez, H. Santos, G. Sanchez, B. Garcia, One possible result is:, President- J. Gubaton, Vice-President- I. Macamay, Secretary- F. Leon, This is different from other possible results, like:, President- H. Santos, Vice-President- R. Cruz, Secretary- J. Gubaton, (Note: It’s hard making a list of all possible, arrangements when the list is long), The number of possible outcomes for the position:, President- 10 possible choices, Vice President- 9 possible choices, Secretary- 8 possible choices, With the use of Fundamental Counting Principle:, , There are 720 possible ways of electing, Mathematics club President, Vice President and, Secretary., Example 4:, Getting the possible, arrangement of, letters that could be, the anagram of the, word BREAK, , 5, , Anagram is a word or phrase created by, rearranging all the letters of a certain word, and, these letters must be used only once., Since we are getting the possible arrangements of, letters of BREAK that has meaning, then the, possible words that could be formed are BRAKE, and BAKER., Thus, there are only 2 possible arrangements., , 5

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Assessment, Guided Assessment, Directions: Write TRUE if the activity/situation illustrates permutation of distinct, objects and write FALSE if it does not., 1. Leg is the anagram of gel., 2. 89659 is a possible PIN of an ATM card., 3. Pagers is a possible arrangement of grapes., 4. The possible arrangement for MANGO is MAGNO., 5. It takes 5040 ways to arrange the letters of JACKETS., 6. Mrs. Dela Cruz arranges 4 the same potted plants in a row., 7. One possible arrangement of STATISTICS is SCITSITATS., 8. Mr. Fernandez hangs 3 the same photo frames in a row on the wall., 9. The possible arrangement of CREAMY when no vowels are allowed is CRMY., 10. It takes 3 ways to arrange 2 identical novel books and 1 algebra book on a shelf., 11. There are 120 possible ways that the 5 people will arrange themselves for picture, taking., 12. There are 504 3-digit numbers can be formed from 1, 2, 3, 4, 5, 6, 7, 8 and 9 when all, digits are distinct., 13. Permutation of distinct objects refers to the different arrangement of distinct objects, in a line., 14. From 8 numbered chips 0, 1, 2, 3, 4, 5, 6, and 7, the 4-digit number with distinct, digits is 1234., 15. The possible PIN for a mobile phone if it is 6-digit even number and digits must be, non-distinct is 600288., , 6

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Lesson, , 2, , Illustrating Permutation of NonDistinct Objects, , What’s In, Permutation refers to the different arrangement of objects in a definitive manner., Permutation of distinct objects is one of the two types of permutation of objects in, a line which refers to the number of different arrangements of different objects in a line. It is, the number of different arrangements when no objects are identical., Examples:, 1. Arranging 3 different mathematics books in a shelf, 2. 67543 is a 6-digit odd number PIN for a mobile phone, 3. Awarding gold, silver and bronze medals to the winning runners in a race, , What’s New, Directions: Rearrange the letters of the following words and find the possible arrangement, of these letters that could be the anagrams of the words below:, 1., 2., 3., 4., 5., , Free, Vases, Peels, Cheaper, Petitioner, , What do you observe of the letters of the words above? Yes, there are duplications of, letters of the words., In this lesson, you will learn about permutation of non-distinct objects. It is another, type of permutation in a line in which it is a permutation when objects are not distinct., Examples:, 1. Arranging the letters of MATHEMATICS, - This permutation is non-distinct because there are duplications of letters. There, are 2 M’s, 2 A’s and 2 T’s., 2. 5 vases of the same kind and 3 candle stands of the same kind are arranged in a line., , 7

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-, , This is permutation of non-distinct objects because there are objects to be, arranged in a line are alike or the same., , What Is It, Let’s find out how to illustrate permutation of non-distinct objects., Situation/ Activity, Example 1:, Finding the number of, possible arrangements of the, letters of the word TREE, , Number of, Objects, , Discussion of the Possible of, Arrangements, , 4, , Since the two E’s are the same, we, can’t make a distinction between the, two. Thus, let us use upper case for, one E and lower case for the other e., The possible arrangements are:, TREe, REet, EeTR ETer, TReE, ReET eETR eTER, TEer, RETe EeRT ERet, TeER, ReTE eERT eRET, TEre, RTEe ETRe ERTe, TeRE, RTeE eTRE eRTE, , Now, these two E’s are the same., Thus, TREe and TReE are also the, same. Therefore, there are 12, possible arrangements of letters of, TREE., We can also use the fundamental, counting principle. Since there are, four letters in the word TREE and, there are two E’s or for each, permutation like T-R-E-E, there are 2, just like what is mentioned above. The, duplications are eliminated by dividing, , Therefore, there are 12 possible, arrangements of letters of TREE., , 8

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Example 2:, Displaying 5 flags in which 3, are red and 2 are yellow, , 5, , Let R for the red flag and Y for the, yellow flag., The possible arrangements of the 5, flags on display are:, RRRYY, RRYYR, RRYRY, RYYRR,, RYRYR, RYRRY, YYRRR, YRRRY,, YRYRR, YRRYR., With the use of fundamental counting, principle: Since there are 5 flags in all, and 3 red flags and 2 yellow flags,, thus, the duplications are eliminated by, dividing, , Therefore, there 10 possible ways of, displaying the 5 flags., Example 3:, Assigning the same feed to 3, pigs and another feed to 3, pigs, , 6, , Suppose these pigs are used in a, study to compare 2 different feeds., Each of these feeds is to be used on 3, randomly selected pigs., Let’s assume that the 2 different feeds, are A and B., The possible ways of feeds to be used, on pigs are:, AAABBB, AABBBA, AABBAB,, AABABB, ABBBAA, ABBAAB,, ABAABB, ABBABA, ABABAB,, ABABBA, BBBAAA, BBAAAB,, BBAABA, BBABAA, BAAABB,, BAABBA, BABBAA, BAABAB,, BABABA, BABAAB., With the use of fundamental counting, principle: Since there are 6pigs in all, and 3 pigs are used for the same feed, and another 3 pigs for another the, same feed, thus, the duplications are, eliminated by dividing, , There are 20 possible ways of feeds to, be used on pigs., , 9

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What’s More, Guided/Controlled Practice, Directions: Write TRUE if the statement is true and write FALSE if otherwise., 1. 10589 is the 5- non-distinct digit PIN of my phone., 2. DIVIDE is the anagram for DIVISION., 3. There are 7 ways to order the word, ARRANGE., 4. The possible arrangement of ADD is DAD., 5. WHEREVER is a permutation of non-distinct objects., 6. EVERYWHERE is not a permutation of non-distinct objects., 7. It takes 100 ways to arrange the letters of the word LETTER., 8. One possible arrangement of MULTIPLICATION is MULTIPLY., 9. There are 12 possible arrangements of letters of the word ROOM., 10. 5 persons are lined up does not involve permutation of non-distinct objects., 11. The objects to be arranged in permutation of non-distinct objects are unique., 12. Getting the anagram of BUTTER is an example of permutation of non-distinct objects., 13. Permutation of non-distinct objects is a different arrangement when objects are not, distinct., 14. You can form 180 different words with or without meaning if you rearrange the letters, of BETTER., 15. Putting 3 plates of the same design and 3 glasses of the same size in a row on the, table involves permutation of non-distinct objects., , What I Can Do, , Assessment, Independent Assessment, Directions: Read each question carefully and choose the letter that corresponds to the, correct answer., 1. Which phrase describes permutation of non-distinct objects?, A. Arrangement of objects in a line, B. Identical objects arranged identically, C. Arrangement of objects that are not distinct, D. Different arrangement of objects that are unique, 2. The following involves permutation of non-distinct objects EXCEPT A. Getting the, arrangement of all letters of REPETITION, B. Placing 3 different blouses and 3 different skirts in a hanging rack, 10

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C. Hanging 4 pieces of wall clock of the same color on the display wall, D. Putting 5 identical plates and 5 different glasses in a row on the table, 3. Which situation illustrates permutation of non-distinct objects?, A. Writing the anagram of builder, B. Unlocking a vault with different digits, C. Having 2 grades of 96 and 3 grades of 91, D. 6 persons sitting around at a circular table, 4. Below are different arrangements of letters, which do you think involves permutation, of non-distinct objects?, A. KEEP, B. QUIET, C. SILENT, D. VOICE, 5. Which arrangement of digits involves permutation of non-distinct objects?, A. 12345, B. 23456, C. 34567, D. 78991, 6. From 6 numbered chips, 0, 1, 2, 3, 4 and 5, which of the following is the 5-digit even, number when digits are non-distinct?, A. 01230, B. 12363, C. 23452, D. 49001, 7. Suppose that the password of a Facebook account consists of 6 characters, which is, the possible password if all characters are non-distinct?, A. Abe_15, B. B@vy14, C. Kiss13, D. M@$_02, 8. Describe the possible arrangement of STREETS., A. RETESTS, B. SETTERS, C. TESTERS, D. all of the above, 9. Which is the possible arrangement that could be the anagram of SPOONS?, A. NOPOSS, B. POSONS, C. SNOOPS, D. SPONOS, 10. What is the possible arrangement of SHEALS which pertain to body part?, A. HASSLE, B. HASSEL, C. LASHES, D. SHALES, 11. The anagram of the word proposes is _____________., A. opposers, B. pospores, C. repospos, D. none of the above, 12. How many ways can a shoemaker display 4 pairs of the same leather shoes, 3 pairs, of the same doll shoes and 2 pairs of different boots in a row in the window?, A. 1260, B. 2520, C. 15120, D. 20160, 13. How many ways can the letters of DIFFERRED be arranged?, A. 1890, B. 7560, 11

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C. 22680, , D. 181440, , 14. Count how many different arrangements can a flower shop owner display 6 vases of, red roses and 4 vases of yellow daisies in a row?, A. 120, B. 210, C. 630, D. 720, 15. How many possible arrangements you can make for hanging 5 plants of the same, kind in a row on a fence?, A. 1, B. 3, C. 5, D. 7, , 12

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Lesson, , 3, , Illustrating Circular, Permutation of Objects, , What Is It, What is circular permutation?, Circular permutation is a type of permutation where the arrangement of different, objects is in circular manner., Examples:, 1. 5 people are seated around a circular table., 2. 5 different keys are arranged in a key ring., 3. 5 different beads on a bracelet, Let’s find out how to illustrate circular permutation of objects., Situation/ Activity, , Number of, Objects, , Discussion of the Possible Arrangements, , 13

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Example 1:, 3 people sitting around a, circular table, , 3, , Suppose the 3 people are A, B and C., Observe the following illustrations., , First, , Fourth, , Second, , Third, , Fifth, , Sixth, , Notice that first, second and third, arrangements are the same. Though the, positions of the guests have been moved but, they are still in the same position in relation to, one another. There is no first place in the, arrangement of guests. When each guest, , 14

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moves his position by one place may it be, clockwise or counterclockwise, it does not, change the relative positions. The same is, true with fourth, fifth and sixth arrangements., These three arrangements as well are the, same. Thus, the three guests can be arranged, around a circular table in 2 ways only., Observe how the same letters A, B and C are, arranged in a line., ABC ACB, CAB BAC, BCA CBA, Notice that the letters can be arranged in 6, ways in a line but the arrangements of them in, circular manner is just 2 ways because the, arrangements falling in each column are just, the same because the 3 guests are supposed, to be seated around a circular table., , With the use of Fundamental Counting, Principle, in getting the number of possible, arrangement of objects in a circular manner is, by just dividing the possible arrangements of, objects in a line by the number of objects, involve in the circular permutation. Like:, , Thus, there are 2 ways that the three guests, can be arranged around a circular table., Example 2:, 4 campers sitting around a, campfire, , 4, , Suppose the campers are A, B, C and D., The possible arrangements of 4 campers are:, , With the use of Fundamental Counting, Principle: Since there are 4 objects involved, then,, , 15

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Therefore, there 6 possible arrangements of 4, campers sitting around a campfire., Example 3:, 8 people sit at a round table, , 8, , Since there 8 people, it’s hard to draw an, illustration of all possible arrangements when, the arrangements are quite long. Then we will, find the number of arrangements by, , Therefore,, there, are, 5040, possible, arrangements that the 8 people sit at a round, table., , Assessment: (Post-Test), Directions: Read each question carefully and choose the letter that corresponds to the, correct answer., 1. It refers to an arrangement of objects in which the order matters., A., Combination, C. Integration, D. Permutation, , B. Differentiation, , 2. The number of ordered arrangements of different objects in a row is called, ___________., A., Permutation, B. Permutation of, distinct objects, C. Circular permutation, D. Permutation of non-distinct objects, 3. An ordered arrangement of objects in a circular manner is called __________., A., Permutation, B. Permutation of, distinct objects, C. Circular permutation, D. Permutation of non-distinct objects, 4. The different arrangement of non-distinct objects in a line is called ____________., A., Circular permutation, B. Permutation of, distinct objects, C. Objective Permutation, D. Permutation of non-distinct objects, 5. Which of the following situations involve permutation?, A., Selecting 3 swimmers from group of 5, B., Choosing friends to exchange greetings, C., Unlocking mobile phone with a passcode, D., Choosing 11 players to play from a soccer team, which has 20 players, 16

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6. The following involve permutations EXCEPT, A., Open a savings account, B., Login to Facebook account, C., Sign into yahoo mail account, D., Enter PIN to login GCASH account, 7. Which of the following situations show that arrangement of the objects is important?, A., Mixing the ingredients of Fruit salad, B., Choosing 3 flavors of ice cream from a menu of, 10, C., Awarding first, second and third place to the, winners, D., Choosing 5 students from a class of 40 to, compete for singing contest, 8. There are 6 ways that 4 persons can be seated at a round table. This is an example, of______________., A., Not permutation, B. Permutation of, distinct objects, C. Circular permutation, D. Permutation of non-distinct objects, 9. Arranging the 10 people in a row is an activity that involves __________., A., Not permutation, B. Permutation of, distinct objects, C. Circular permutation, D. Permutation of non-distinct objects, 10. A bag shop owner wants to display her new collection of bags in the front window., She displays 2 identical Gucci bags, 2 identical Hermes bags and 2 identical LV, bags. This problem is _____________________., A., Not permutation, B. Permutation of, distinct objects, C. Circular permutation, D. Permutation of non-distinct objects, 11. The following are possible 5 odd non-distinct digit PIN’s of the computer, EXCEPT_________., A., 11331, B. 23431, C. 34565, D. 45679, 12. Given the 7-letter word honesty, which of the following possible arrangement of the, letters, 4 letters at a time?, A., Hone, B. Ones, C. Nest, D. all of the above, 13. What is the possible arrangement of 5 persons (3 boys and 2 girls) sitting around a, circular table if 2 girls sit beside each, other?, A., B., , C., , D., 17

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14. How many possible arrangements of 10 persons sitting around a circular table?, A., 3024, B. 15120, C. 60480, D. 362880, 15. How many possible arrangements you will made in arranging the word, TESTS?, A., 10, B. 20, C. 30, D. 40, , 18