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G10M-Q3-L1-Permutation-of-Objects-Grade 10.pptx

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Permutation of Objects for Grade 10

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PERMUTATION OF OBJECTS
SITUATION: Teacher Sarah accepts tutorial services in her house every Saturday. She wanted to check if the seating arrangement of her three tutees (Ana, Beauty and Carl) has a significant difference in their learning. So, she decided to change their seating arrangement every Saturday.
PERMUTATION OF OBJECTS Permutation is an arrangement of all or part of a set of objects with proper regard to order. In short, ORDER MATTERS. Examples: β€’ Possible number of positions in a picture frame β€’ Possible number of plate number LTO can create β€’ Possible gender of the children a couple could have β€’ Possible choices you have in creating your password β€’ Possible arrangements of books in a shelf
ACTIVITY 1.1: PERMUTATION OR NOT Tell whether each of the problems fall under permutation or not. 5. Voting the President, VP and Secretary from a class 4. Picking a team of 3 from a group of 10 2. Choosing 3 dishes in a buffet 1. Listing 5 games for a party 3. Placing winners in a running race β†’ permutation β†’ permutation β†’ not permutation β†’ not permutation β†’ permutation
Fundamental Counting Principle β†’ a rule used to count the total number of possible outcomes in a situation. It states that if there are 𝒏 ways of doing something, and π’Ž ways of doing another thing after that, then there are 𝒏 Γ— π’Ž ways to perform both of these actions. METHODS IN ILLUSTRATING AND DETERMINING THE PERMUTATION OF OBJECTS Systematic Listing β†’ involves coming up with an actual list of all possible outcomes in an organized and systematic way. Tree Diagram β†’ display all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. Table β†’ displays all the possible outcomes of an event in a tabular way.
ACTIVITY 1.2: ARRANGE THE SEATS Teacher Sarah accepts tutorial services in her house every Saturday. She wanted to check if the seating arrangement of her three tutees (Ana, Beauty and Carl) has a significant difference in their learning. So, she decided to change their seating arrangement every Saturday.
ACTIVITY 1.2: ARRANGE THE SEATS Answer the following questions a. List all possible seating arrangements of the three tutees? 1 β†’ 2 β†’ 3 β†’ 4 β†’ 5 β†’ 6 β†’ b. In how many possible ways can they be seated in a row? c. What method is use to find the possible arrangements? β†’ 6 possible ways β†’ Systematic listing
ACTIVITY 1.2: ARRANGE THE SEATS d. Show another way to find the possible arrangements. 1 (Ana, Beauty, Carl) 2 (Ana, Carl, Beauty) 3 (Beauty, Ana, Carl) 4 (Beauty, Carl, Ana) 5 (Carl, Ana, Beauty) 6 (Carl, Beauty, Ana) 1st 2nd 3rd 1st 2nd 3rd 1st 2nd 3rd β†’ By using tree diagram
ACTIVITY 1.2: ARRANGE THE SEATS d. Show another way to find the possible arrangements. 1 2 3 4 5 6 1st chair 2nd chair 3rd chair Ana Beauty Carl Ana Carl Beauty Beauty Ana Carl Beauty Carl Ana Carl Ana Beauty Carl Beauty Ana β†’ By using a table
ACTIVITY 1.2: ARRANGE THE SEATS e. Show another way to find the number of possible arrangements. Tutees: Ana, Beauty, and Carl Chairs: 1st, 2nd, and 3rd Γ— Γ— 3 2 1 = 6 possible arrangements 1st 2nd 3rd β†’ Use the fundamental counting principle
ACTIVITY 1.3: BREAK THE CODE Suppose you secure your cellphone using a 4-digit code. You forgot the code and only remembered that the code contains the digits 2, 3, 4, and 5. List all the possible codes out of the given digits. Determine the number of permutations made. 2, 3, 4, 5 2, 3, 5, 4 2, 4, 3, 5 2, 4, 5, 3 2, 5, 3, 4 2, 5, 4, 3 3, 2, 4, 5 3, 2, 5, 4 3, 4, 2, 5 3, 4, 5, 2 3, 5, 2, 4 3, 5, 4, 2 4, 2, 3, 5 4, 2, 5, 3 4, 3, 2, 5 4, 3, 5, 2 4, 5, 2, 3 4, 5, 3, 2 5, 2, 3, 4 5, 2, 4, 3 5, 3, 2, 4 5, 3, 4, 2 5, 4, 2, 3 5, 4, 3, 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 24 possible codes
ACTIVITY 1.3: BREAK THE CODE Suppose you secure your cellphone using a 4-digit code. You forgot the code and only remembered that the code contains the digits 2, 3, 4, and 5. List all the possible codes out of the given digits. Determine the number of permutations made. Digits: 2, 3, 4, and 5 Code: 4 digits Γ— Γ— 3 2 1 = 24 possible ways 2nd 3rd 4th Check using the fundamental counting principle Γ— 4 1st 2 3 4 5 3 4 5 4 5 5
ACTIVITY 1.4: SON OR DAUGHTER? A couple is planning to have 4 kids. On any birth the children could be a son or a daughter. Use a tree diagram to illustrate the permutations. 1 (S, S, S, S) 1st 2nd 3rd D S S D D S S D D S D S S D S 4th 1st 2nd 3rd D S S D D S S D D S D S S D D 4th 2 (S, S, S, D) 3 (S, S, D, S) 4 (S, S, D, D) 5 (S, D, S, S) 6 (S, D, S, D) 7 (S, D, D, S) 8 (S, D, D, D) 9 (D, S, S, S) 10 (D, S, S, D) 11 (D, S, D, S) 12 (D, S, D, D) 13 (D, D, S, S) 14 (D, D, S, D) 15 (D, D, D, S) 16 (D, D, D, D) 16 possible permutations
ACTIVITY 1.4: SON OR DAUGHTER? A couple is planning to have 4 kids. On any birth the children could be a son of a daughter. Use a tree diagram to illustrate the permutations. Kids: Son or Daughter Number of kids: 4 Γ— Γ— 2 2 2 = 16 possible permutations 2nd 3rd 4th Check using the fundamental counting principle Γ— 2 1st S D S D S D S D
ACTIVITY 1.5: LET’S TRY IT TOGETHER Rosa was invited to a birthday party. She is planning to match a blouse and a skirt to wear. She has 4 new blouses (stripes, with ruffles, long- sleeved, and sleeveless) and 3 skirts (red, pink, and black) in her closet for such occasions. Make a table to show the possible outfit she can wear. How many ways can she select her outfit? 1 BLOUSE SKIRT stripes red stripes pink stripes black with ruffles red with ruffles pink with ruffles black long-sleeved red long-sleeved pink long-sleeved black sleeveless red sleeveless pink sleeveless black 2 3 4 5 6 7 8 9 10 11 12 β†’ 12 possible outfits
ACTIVITY 1.5: LET’S TRY IT TOGETHER Rosa was invited to a birthday party. She is planning to match a blouse and a skirt to wear. She has 4 new blouses (stripes, with ruffles, long- sleeved, and sleeveless) and 3 skirts (red, pink, and black) in her closet for such occasions. Make a table to show the possible outfit she can wear. How many ways can she select her outfit? β†’ 12 possible outfits 3 2nd Γ— 4 1st R P B WR LS SL Blouses: stripes (ST), with ruffles (WR), long-sleeved (LS), sleeveless (LS) Skirts: red (R), pink (P), and black (B) Check using the fundamental counting principle ST = 12 possible outfits
EVALUATION 1.1: LET’S TRY IT INDIVIDUALLY In how many ways can you place 6 different books on a shelf? 6 different books Γ— Γ— 5 4 3 = 720 possible ways 2nd 3rd 4th Given: Γ— 6 1st 6 books 5 books 4 books 3 books Γ— 2 5th 2 books Γ— 1 6th 1 book
ACTIVITY 1.6: HOW MANY WAYS? In how many ways can you arrange 5 students in a row? 5 students Γ— Γ— 5 4 3 = 120 possible ways 1st 2nd 3rd Given: 5 students 4 students 3 students Γ— 2 4th 2 students Γ— 1 5th 1 student
FACTORIAL NOTATION Factorial Notation is the product of all positive integers less than or equal to 𝒏. It is denoted by 𝒏! Example: 5! 5! can be read as β€œ5 factorial”. 5! means 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 = 120. 5! can be written in many ways. 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 5 Γ— 4 Γ— 3 Γ— 2! 5 Γ— 4 Γ— 3! 5 Γ— 4! 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 5 βˆ™ 4 βˆ™ 3 βˆ™ 2! 5 βˆ™ 4 βˆ™ 3! 5 βˆ™ 4!
FACTORIAL NOTATION 𝒏 𝑬𝒙𝒑𝒂𝒏𝒅𝒆𝒅 π‘­π’π’“π’Ž 𝑹𝒆𝒔𝒖𝒍𝒕 𝟎! 𝟏! 𝟐! πŸ‘! πŸ’! πŸ“! πŸ”! πŸ•! πŸ–! πŸ—! πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ‘ Γ— 𝟐 Γ— 𝟏 𝟐 Γ— 𝟏 𝟏 𝟏 πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ• Γ— πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ– Γ— πŸ• Γ— πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ— Γ— πŸ– Γ— πŸ• Γ— πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 𝟏 𝟏 𝟐 πŸ” πŸπŸ’ 𝟏𝟐𝟎 πŸ•πŸπŸŽ πŸ“, πŸŽπŸ’πŸŽ πŸ’πŸŽ, πŸ‘πŸπŸŽ πŸ‘πŸ”πŸ, πŸ–πŸ–πŸŽ
ACTIVITY 1.7: EVALUATE THE FACTORIAL Evaluate the following expressions. 1. 5! + 2! = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 + 2 βˆ™ 1 = 120 + 2 = 𝟏𝟐𝟐 2. 8! βˆ’ 5! = 8 βˆ™ 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 40, 320 βˆ’ 120 = πŸ’πŸŽ, 𝟐𝟎𝟎 βˆ’ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 3. 4! 2! = 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 2 βˆ™ 1 = (24)(2) = πŸ’πŸ–
ACTIVITY 1.8: LET’S TRY IT TOGETHER Evaluate the following expressions. 4. 6! 3! = 𝟏𝟐𝟎 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 3 βˆ™ 2 βˆ™ 1 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3! 3! = 𝟏𝟐𝟎 5. 7! (7 βˆ’ 2)! = πŸ’πŸ = 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 7! 5! = 7 βˆ™ 6 βˆ™ 5! 5! = πŸ’πŸ 6. 5! 3! 2! = 5 βˆ™ 4 2 βˆ™ 1 = 5 βˆ™ 4 βˆ™ 3! 3! 2! = 20 2 = 𝟏𝟎
EVALUATION 1.2: LET’S TRY IT INDIVIDUALLY Determine whether each equation is True of False 1. 5! = 5 βˆ™ 4 βˆ™ 3! 2. 3! + 3! = 6! 3. 4! 5! = 9! 5. 10! 7! = 10 βˆ™ 9 βˆ™ 8 6. 5! 3! 2! = 5 βˆ™ 2! 4. 5 βˆ’ 3 ! = 5! βˆ’ 3!
EVALUATION 1.2: LET’S TRY IT INDIVIDUALLY Determine whether each equation is True of False 1. 5! = 5 βˆ™ 4 βˆ™ 3! 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 120 = 120 β†’ True 2. 3! + 3! = 6! 3. 4! 5! = 9! 3 βˆ™ 2 βˆ™ 1 + 3 βˆ™ 2 βˆ™ 1 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 12 = 720 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 9 βˆ™ 8 βˆ™ 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 2, 880 = 362, 880 β†’ False β†’ False
EVALUATION 1.2: LET’S TRY IT INDIVIDUALLY Determine whether each equation is True of False β†’ True 5. 10! 7! = 10 βˆ™ 9 βˆ™ 8 6. 5! 3! 2! = 5 βˆ™ 2! 10 βˆ™ 9 βˆ™ 8 βˆ™ 7! 7! = 10 βˆ™ 9 βˆ™ 8 10 βˆ™ 9 βˆ™ 8 = 10 βˆ™ 9 βˆ™ 8 720 = 720 5 βˆ™ 4 βˆ™ 3! 3! 2! = 10 5 βˆ™ 4 2 βˆ™ 1 = 10 20 2 = 10 10 = 10 β†’ True 4. 5 βˆ’ 3 ! = 5! βˆ’ 3! 2! = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 βˆ’ (3 βˆ™ 2 βˆ™ 1) 2 βˆ™ 1 = 120 βˆ’ 6 2 = 114 β†’ False
ACTIVITY 1.9: LET’S FIND OUT! Answer the following questions. b. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 5-digit passcode if repetition is not allowed? a. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 5-digit passcode if repetition is allowed? 5 digits 5-digit passcode Γ— 5 5 = πŸ‘πŸπŸπŸ“ possible ways Given: Γ— 5 Γ— 5 5 Γ— 5 digits 5-digit passcode Γ— 4 3 = 𝟏𝟐𝟎 possible ways Given: Γ— 5 Γ— 2 1 Γ—
ACTIVITY 1.9: LET’S FIND OUT! Answer the following questions. d. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 3-digit passcode if repetition is not allowed? c. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 3-digit passcode if repetition is allowed? 5 digits 3-digit passcode Γ— 5 5 = 125 possible ways Given: Γ— 5 5 digits 3-digit passcode Γ— 4 3 = πŸ”πŸŽ possible ways Given: Γ— 5
DERIVATION OF PERMUTATION FORMULA A. When repetition is allowed: P(5 π‘‘π‘Žπ‘˜π‘’π‘› 5) = 5 βˆ™ 5 βˆ™ 5 βˆ™ 5 βˆ™ 5 = 55 = 𝒏𝒓 P(5 π‘‘π‘Žπ‘˜π‘’π‘› 3) = 5 βˆ™ 5 βˆ™ 5 = 53 = 𝒏𝒓 It is important in solving permutation to consider these questions: a) Is repetition allowed? b) Is repetition not allowed?
DERIVATION OF PERMUTATION FORMULA P(5 π‘‘π‘Žπ‘˜π‘’π‘› 5) = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 P(5 π‘‘π‘Žπ‘˜π‘’π‘› 3) = 5 βˆ™ 4 βˆ™ 3 = 5! = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 2 βˆ™ 1 = 5! 2! = 5! (5 βˆ’ 3)! = 𝒏! = 𝒏! (𝒏 βˆ’ 𝒓)! (5 βˆ’ 3) B. When repetition is not allowed:
A. PERMUTATION WHEN REPETITION IS ALLOWED The number of permutations of n objects when repetition is allowed is determined by the formula: where 𝒏 is the number of objects and 𝒓 is the number of objects selected. 𝒏𝑷𝒓 = 𝒏𝒓 b. Permutation of All Objects Taken r at a Time The number of permutations of n objects taken r at a time is determined by the formula: 𝒏𝑷𝒓 = 𝒏! (π’βˆ’π’“)! where 𝒏 is the number of objects and 𝒓 is the number of objects selected. a. Permutation Of All Objects Taken All at a Time The number of permutations of n objects taken all at a time is determined by the formula: 𝒏𝑷𝒏 = 𝒏! where 𝒏 is the number of objects. B. PERMUTATION WHEN REPETITION IS NOT ALLOWED
ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 3. In how many ways can 3 runners finished the race if there are no ties? 2. In how many ways are there to place four different colored tiles in a row? 4. In how many ways can 7 different books can be arranged on a shelf if there is space enough for 4 books? 5. In how many ways can a president and vice president be chosen from a club with 10 members? 1. In how many ways are there to arrange the letters ABCD if repetition is allowed?
ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 1. In how many ways are there to arrange the letters ABCD if repetition is allowed? 4 letters (All are taken and repetition is allowed) Given: 𝒏𝑷𝒓 = 𝒏𝒓 4𝑃4 = πŸ’πŸ’ 4𝑃4 = 4 βˆ™ 4 βˆ™ 4 βˆ™ 4 πŸ’π‘·πŸ’ = πŸπŸ“πŸ” π’˜π’‚π’šπ’” 𝑷(πŸ’, πŸ’) = πŸπŸ“πŸ” π’˜π’‚π’šπ’”
ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 2. In how many ways are there to place four different colored tiles in a row? 4 colored tiles (All are taken and repetition is not allowed) Given: 𝒏𝑷𝒏 = 𝒏! 4𝑃4 = 4! 4𝑃4 = 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 πŸ’π‘·πŸ’ = πŸπŸ’ π’˜π’‚π’šπ’” 𝑷(πŸ’, πŸ’) = πŸπŸ’ π’˜π’‚π’šπ’”
ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 3. In how many ways can 3 runners finished the race if there are no ties? 3 runners (All are taken and repetition is not allowed) Given: 𝒏𝑷𝒏 = 𝒏! 3𝑃3 = 3! 3𝑃3 = 3 βˆ™ 2 βˆ™ 1 πŸ‘π‘·πŸ‘ = πŸ” π’˜π’‚π’šπ’” 𝑷(πŸ‘, πŸ‘) = πŸ” π’˜π’‚π’šπ’”
ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 4. In how many ways can 7 different books can be arranged on a shelf if there is space enough for 4 books? 7 different books (4 are taken and repetition is not allowed) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 7𝑃4 = 7! (7 βˆ’ 4)! 7𝑃4 = 7! 3! 7𝑃4 = 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3! 3! 7𝑃4 = 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 πŸ•π‘·πŸ’ = πŸ–πŸ’πŸŽ π’˜π’‚π’šπ’” 𝑷(πŸ•, πŸ’) = πŸ–πŸ’πŸŽ π’˜π’‚π’šπ’”
ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 5. In how many ways can a president and vice president be chosen from a club with 6 members? 6 members (2 are taken and repetition is not allowed) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 6𝑃2 = 6! (6 βˆ’ 2)! 6𝑃2 = 6! 4! 6𝑃2 = 6 βˆ™ 5 βˆ™ 4! 4! 6𝑃2 = 6 βˆ™ 5 πŸ”π‘·πŸ = πŸ‘πŸŽ π’˜π’‚π’šπ’” 𝑷(πŸ”, 𝟐) = πŸ‘πŸŽ π’˜π’‚π’šπ’”
EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 2. In how many ways can the digits 1, 2, 3, 4, 5, and 6 be arranged to make a 3-digit code if repetition is allowed? 1. In how many ways can you arrange 8 figurines on a shelf? 3. In how many ways can 3 bicycles be parked if there are 5 available parking spaces? 4. In how many ways can Aling Maria arrange her 6 different potted plants in her plant stand if only 3 plants can fit?
EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 1. In how many ways can you arrange 8 figurines on a shelf? 8 figurines (All are taken and repetition is not allowed) Given: 𝒏𝑷𝒏 = 𝒏! 8𝑃8 = 8! 8𝑃8 = 8 βˆ™ 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 πŸ–π‘·πŸ– = πŸ’πŸŽ, πŸ‘πŸπŸŽ π’˜π’‚π’šπ’” 𝑷 πŸ’, πŸ’ = πŸ’πŸŽ, πŸ‘πŸπŸŽ π’˜π’‚π’šπ’”
EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 2. In how many ways can the digits 1, 2, 3, 4, 5, and 6 be arranged to make a 3-digit code if repetition is allowed? 6 digits (3 are taken and repetition is allowed) Given: 𝒏𝑷𝒓 = 𝒏𝒓 6𝑃3 = 63 6𝑃3 = 6 βˆ™ 6 βˆ™ 6 πŸ”π‘·πŸ‘ = πŸπŸπŸ” π’˜π’‚π’šπ’” 𝑷(πŸ”, πŸ‘) = πŸπŸπŸ” π’˜π’‚π’šπ’”
EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 3. In how many ways can 3 bicycles be parked if there are 5 available parking spaces? 5 parking spaces (3 are taken and repetition is not allowed)) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 5𝑃3 = 5! (5 βˆ’ 3)! 5𝑃3 = 5! 2! 5𝑃3 = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2! 2! 5𝑃3 = 5 βˆ™ 4 βˆ™ 3 πŸ“π‘·πŸ‘ = πŸ”πŸŽ π’˜π’‚π’šπ’” 𝑷(πŸ“, πŸ‘) = πŸ”πŸŽ π’˜π’‚π’šπ’”
EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 4. In how many ways can Aling Maria arrange her 6 different potted plants in her plant stand if only 3 plants can fit? 6 potted plants (3 are taken and repetition is not allowed)) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 6𝑃3 = 6! (6 βˆ’ 3)! 6𝑃3 = 6! 3! 6𝑃3 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3! 3! 6𝑃3 = 6 βˆ™ 5 βˆ™ 4 πŸ”π‘·πŸ‘ = 𝟏𝟐𝟎 π’˜π’‚π’šπ’” 𝑷(πŸ”, πŸ‘) = 𝟏𝟐𝟎 π’˜π’‚π’šπ’”

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G10M-Q3-L1-Permutation-of-Objects-Grade 10.pptx

  • 2. SITUATION: Teacher Sarah accepts tutorial services in her house every Saturday. She wanted to check if the seating arrangement of her three tutees (Ana, Beauty and Carl) has a significant difference in their learning. So, she decided to change their seating arrangement every Saturday.
  • 3. PERMUTATION OF OBJECTS Permutation is an arrangement of all or part of a set of objects with proper regard to order. In short, ORDER MATTERS. Examples: β€’ Possible number of positions in a picture frame β€’ Possible number of plate number LTO can create β€’ Possible gender of the children a couple could have β€’ Possible choices you have in creating your password β€’ Possible arrangements of books in a shelf
  • 4. ACTIVITY 1.1: PERMUTATION OR NOT Tell whether each of the problems fall under permutation or not. 5. Voting the President, VP and Secretary from a class 4. Picking a team of 3 from a group of 10 2. Choosing 3 dishes in a buffet 1. Listing 5 games for a party 3. Placing winners in a running race β†’ permutation β†’ permutation β†’ not permutation β†’ not permutation β†’ permutation
  • 5. Fundamental Counting Principle β†’ a rule used to count the total number of possible outcomes in a situation. It states that if there are 𝒏 ways of doing something, and π’Ž ways of doing another thing after that, then there are 𝒏 Γ— π’Ž ways to perform both of these actions. METHODS IN ILLUSTRATING AND DETERMINING THE PERMUTATION OF OBJECTS Systematic Listing β†’ involves coming up with an actual list of all possible outcomes in an organized and systematic way. Tree Diagram β†’ display all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. Table β†’ displays all the possible outcomes of an event in a tabular way.
  • 6. ACTIVITY 1.2: ARRANGE THE SEATS Teacher Sarah accepts tutorial services in her house every Saturday. She wanted to check if the seating arrangement of her three tutees (Ana, Beauty and Carl) has a significant difference in their learning. So, she decided to change their seating arrangement every Saturday.
  • 7. ACTIVITY 1.2: ARRANGE THE SEATS Answer the following questions a. List all possible seating arrangements of the three tutees? 1 β†’ 2 β†’ 3 β†’ 4 β†’ 5 β†’ 6 β†’ b. In how many possible ways can they be seated in a row? c. What method is use to find the possible arrangements? β†’ 6 possible ways β†’ Systematic listing
  • 8. ACTIVITY 1.2: ARRANGE THE SEATS d. Show another way to find the possible arrangements. 1 (Ana, Beauty, Carl) 2 (Ana, Carl, Beauty) 3 (Beauty, Ana, Carl) 4 (Beauty, Carl, Ana) 5 (Carl, Ana, Beauty) 6 (Carl, Beauty, Ana) 1st 2nd 3rd 1st 2nd 3rd 1st 2nd 3rd β†’ By using tree diagram
  • 9. ACTIVITY 1.2: ARRANGE THE SEATS d. Show another way to find the possible arrangements. 1 2 3 4 5 6 1st chair 2nd chair 3rd chair Ana Beauty Carl Ana Carl Beauty Beauty Ana Carl Beauty Carl Ana Carl Ana Beauty Carl Beauty Ana β†’ By using a table
  • 10. ACTIVITY 1.2: ARRANGE THE SEATS e. Show another way to find the number of possible arrangements. Tutees: Ana, Beauty, and Carl Chairs: 1st, 2nd, and 3rd Γ— Γ— 3 2 1 = 6 possible arrangements 1st 2nd 3rd β†’ Use the fundamental counting principle
  • 11. ACTIVITY 1.3: BREAK THE CODE Suppose you secure your cellphone using a 4-digit code. You forgot the code and only remembered that the code contains the digits 2, 3, 4, and 5. List all the possible codes out of the given digits. Determine the number of permutations made. 2, 3, 4, 5 2, 3, 5, 4 2, 4, 3, 5 2, 4, 5, 3 2, 5, 3, 4 2, 5, 4, 3 3, 2, 4, 5 3, 2, 5, 4 3, 4, 2, 5 3, 4, 5, 2 3, 5, 2, 4 3, 5, 4, 2 4, 2, 3, 5 4, 2, 5, 3 4, 3, 2, 5 4, 3, 5, 2 4, 5, 2, 3 4, 5, 3, 2 5, 2, 3, 4 5, 2, 4, 3 5, 3, 2, 4 5, 3, 4, 2 5, 4, 2, 3 5, 4, 3, 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 24 possible codes
  • 12. ACTIVITY 1.3: BREAK THE CODE Suppose you secure your cellphone using a 4-digit code. You forgot the code and only remembered that the code contains the digits 2, 3, 4, and 5. List all the possible codes out of the given digits. Determine the number of permutations made. Digits: 2, 3, 4, and 5 Code: 4 digits Γ— Γ— 3 2 1 = 24 possible ways 2nd 3rd 4th Check using the fundamental counting principle Γ— 4 1st 2 3 4 5 3 4 5 4 5 5
  • 13. ACTIVITY 1.4: SON OR DAUGHTER? A couple is planning to have 4 kids. On any birth the children could be a son or a daughter. Use a tree diagram to illustrate the permutations. 1 (S, S, S, S) 1st 2nd 3rd D S S D D S S D D S D S S D S 4th 1st 2nd 3rd D S S D D S S D D S D S S D D 4th 2 (S, S, S, D) 3 (S, S, D, S) 4 (S, S, D, D) 5 (S, D, S, S) 6 (S, D, S, D) 7 (S, D, D, S) 8 (S, D, D, D) 9 (D, S, S, S) 10 (D, S, S, D) 11 (D, S, D, S) 12 (D, S, D, D) 13 (D, D, S, S) 14 (D, D, S, D) 15 (D, D, D, S) 16 (D, D, D, D) 16 possible permutations
  • 14. ACTIVITY 1.4: SON OR DAUGHTER? A couple is planning to have 4 kids. On any birth the children could be a son of a daughter. Use a tree diagram to illustrate the permutations. Kids: Son or Daughter Number of kids: 4 Γ— Γ— 2 2 2 = 16 possible permutations 2nd 3rd 4th Check using the fundamental counting principle Γ— 2 1st S D S D S D S D
  • 15. ACTIVITY 1.5: LET’S TRY IT TOGETHER Rosa was invited to a birthday party. She is planning to match a blouse and a skirt to wear. She has 4 new blouses (stripes, with ruffles, long- sleeved, and sleeveless) and 3 skirts (red, pink, and black) in her closet for such occasions. Make a table to show the possible outfit she can wear. How many ways can she select her outfit? 1 BLOUSE SKIRT stripes red stripes pink stripes black with ruffles red with ruffles pink with ruffles black long-sleeved red long-sleeved pink long-sleeved black sleeveless red sleeveless pink sleeveless black 2 3 4 5 6 7 8 9 10 11 12 β†’ 12 possible outfits
  • 16. ACTIVITY 1.5: LET’S TRY IT TOGETHER Rosa was invited to a birthday party. She is planning to match a blouse and a skirt to wear. She has 4 new blouses (stripes, with ruffles, long- sleeved, and sleeveless) and 3 skirts (red, pink, and black) in her closet for such occasions. Make a table to show the possible outfit she can wear. How many ways can she select her outfit? β†’ 12 possible outfits 3 2nd Γ— 4 1st R P B WR LS SL Blouses: stripes (ST), with ruffles (WR), long-sleeved (LS), sleeveless (LS) Skirts: red (R), pink (P), and black (B) Check using the fundamental counting principle ST = 12 possible outfits
  • 17. EVALUATION 1.1: LET’S TRY IT INDIVIDUALLY In how many ways can you place 6 different books on a shelf? 6 different books Γ— Γ— 5 4 3 = 720 possible ways 2nd 3rd 4th Given: Γ— 6 1st 6 books 5 books 4 books 3 books Γ— 2 5th 2 books Γ— 1 6th 1 book
  • 18. ACTIVITY 1.6: HOW MANY WAYS? In how many ways can you arrange 5 students in a row? 5 students Γ— Γ— 5 4 3 = 120 possible ways 1st 2nd 3rd Given: 5 students 4 students 3 students Γ— 2 4th 2 students Γ— 1 5th 1 student
  • 19. FACTORIAL NOTATION Factorial Notation is the product of all positive integers less than or equal to 𝒏. It is denoted by 𝒏! Example: 5! 5! can be read as β€œ5 factorial”. 5! means 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 = 120. 5! can be written in many ways. 5 Γ— 4 Γ— 3 Γ— 2 Γ— 1 5 Γ— 4 Γ— 3 Γ— 2! 5 Γ— 4 Γ— 3! 5 Γ— 4! 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 5 βˆ™ 4 βˆ™ 3 βˆ™ 2! 5 βˆ™ 4 βˆ™ 3! 5 βˆ™ 4!
  • 20. FACTORIAL NOTATION 𝒏 𝑬𝒙𝒑𝒂𝒏𝒅𝒆𝒅 π‘­π’π’“π’Ž 𝑹𝒆𝒔𝒖𝒍𝒕 𝟎! 𝟏! 𝟐! πŸ‘! πŸ’! πŸ“! πŸ”! πŸ•! πŸ–! πŸ—! πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ‘ Γ— 𝟐 Γ— 𝟏 𝟐 Γ— 𝟏 𝟏 𝟏 πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ• Γ— πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ– Γ— πŸ• Γ— πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 πŸ— Γ— πŸ– Γ— πŸ• Γ— πŸ” Γ— πŸ“ Γ— πŸ’ Γ— πŸ‘ Γ— 𝟐 Γ— 𝟏 𝟏 𝟏 𝟐 πŸ” πŸπŸ’ 𝟏𝟐𝟎 πŸ•πŸπŸŽ πŸ“, πŸŽπŸ’πŸŽ πŸ’πŸŽ, πŸ‘πŸπŸŽ πŸ‘πŸ”πŸ, πŸ–πŸ–πŸŽ
  • 21. ACTIVITY 1.7: EVALUATE THE FACTORIAL Evaluate the following expressions. 1. 5! + 2! = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 + 2 βˆ™ 1 = 120 + 2 = 𝟏𝟐𝟐 2. 8! βˆ’ 5! = 8 βˆ™ 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 40, 320 βˆ’ 120 = πŸ’πŸŽ, 𝟐𝟎𝟎 βˆ’ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 3. 4! 2! = 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 2 βˆ™ 1 = (24)(2) = πŸ’πŸ–
  • 22. ACTIVITY 1.8: LET’S TRY IT TOGETHER Evaluate the following expressions. 4. 6! 3! = 𝟏𝟐𝟎 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 3 βˆ™ 2 βˆ™ 1 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3! 3! = 𝟏𝟐𝟎 5. 7! (7 βˆ’ 2)! = πŸ’πŸ = 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 7! 5! = 7 βˆ™ 6 βˆ™ 5! 5! = πŸ’πŸ 6. 5! 3! 2! = 5 βˆ™ 4 2 βˆ™ 1 = 5 βˆ™ 4 βˆ™ 3! 3! 2! = 20 2 = 𝟏𝟎
  • 23. EVALUATION 1.2: LET’S TRY IT INDIVIDUALLY Determine whether each equation is True of False 1. 5! = 5 βˆ™ 4 βˆ™ 3! 2. 3! + 3! = 6! 3. 4! 5! = 9! 5. 10! 7! = 10 βˆ™ 9 βˆ™ 8 6. 5! 3! 2! = 5 βˆ™ 2! 4. 5 βˆ’ 3 ! = 5! βˆ’ 3!
  • 24. EVALUATION 1.2: LET’S TRY IT INDIVIDUALLY Determine whether each equation is True of False 1. 5! = 5 βˆ™ 4 βˆ™ 3! 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 120 = 120 β†’ True 2. 3! + 3! = 6! 3. 4! 5! = 9! 3 βˆ™ 2 βˆ™ 1 + 3 βˆ™ 2 βˆ™ 1 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 12 = 720 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 = 9 βˆ™ 8 βˆ™ 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 2, 880 = 362, 880 β†’ False β†’ False
  • 25. EVALUATION 1.2: LET’S TRY IT INDIVIDUALLY Determine whether each equation is True of False β†’ True 5. 10! 7! = 10 βˆ™ 9 βˆ™ 8 6. 5! 3! 2! = 5 βˆ™ 2! 10 βˆ™ 9 βˆ™ 8 βˆ™ 7! 7! = 10 βˆ™ 9 βˆ™ 8 10 βˆ™ 9 βˆ™ 8 = 10 βˆ™ 9 βˆ™ 8 720 = 720 5 βˆ™ 4 βˆ™ 3! 3! 2! = 10 5 βˆ™ 4 2 βˆ™ 1 = 10 20 2 = 10 10 = 10 β†’ True 4. 5 βˆ’ 3 ! = 5! βˆ’ 3! 2! = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 βˆ’ (3 βˆ™ 2 βˆ™ 1) 2 βˆ™ 1 = 120 βˆ’ 6 2 = 114 β†’ False
  • 26. ACTIVITY 1.9: LET’S FIND OUT! Answer the following questions. b. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 5-digit passcode if repetition is not allowed? a. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 5-digit passcode if repetition is allowed? 5 digits 5-digit passcode Γ— 5 5 = πŸ‘πŸπŸπŸ“ possible ways Given: Γ— 5 Γ— 5 5 Γ— 5 digits 5-digit passcode Γ— 4 3 = 𝟏𝟐𝟎 possible ways Given: Γ— 5 Γ— 2 1 Γ—
  • 27. ACTIVITY 1.9: LET’S FIND OUT! Answer the following questions. d. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 3-digit passcode if repetition is not allowed? c. In how many ways the digits 1,2,3,4, and 5 be arranged to make a 3-digit passcode if repetition is allowed? 5 digits 3-digit passcode Γ— 5 5 = 125 possible ways Given: Γ— 5 5 digits 3-digit passcode Γ— 4 3 = πŸ”πŸŽ possible ways Given: Γ— 5
  • 28. DERIVATION OF PERMUTATION FORMULA A. When repetition is allowed: P(5 π‘‘π‘Žπ‘˜π‘’π‘› 5) = 5 βˆ™ 5 βˆ™ 5 βˆ™ 5 βˆ™ 5 = 55 = 𝒏𝒓 P(5 π‘‘π‘Žπ‘˜π‘’π‘› 3) = 5 βˆ™ 5 βˆ™ 5 = 53 = 𝒏𝒓 It is important in solving permutation to consider these questions: a) Is repetition allowed? b) Is repetition not allowed?
  • 29. DERIVATION OF PERMUTATION FORMULA P(5 π‘‘π‘Žπ‘˜π‘’π‘› 5) = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 P(5 π‘‘π‘Žπ‘˜π‘’π‘› 3) = 5 βˆ™ 4 βˆ™ 3 = 5! = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 2 βˆ™ 1 = 5! 2! = 5! (5 βˆ’ 3)! = 𝒏! = 𝒏! (𝒏 βˆ’ 𝒓)! (5 βˆ’ 3) B. When repetition is not allowed:
  • 30. A. PERMUTATION WHEN REPETITION IS ALLOWED The number of permutations of n objects when repetition is allowed is determined by the formula: where 𝒏 is the number of objects and 𝒓 is the number of objects selected. 𝒏𝑷𝒓 = 𝒏𝒓 b. Permutation of All Objects Taken r at a Time The number of permutations of n objects taken r at a time is determined by the formula: 𝒏𝑷𝒓 = 𝒏! (π’βˆ’π’“)! where 𝒏 is the number of objects and 𝒓 is the number of objects selected. a. Permutation Of All Objects Taken All at a Time The number of permutations of n objects taken all at a time is determined by the formula: 𝒏𝑷𝒏 = 𝒏! where 𝒏 is the number of objects. B. PERMUTATION WHEN REPETITION IS NOT ALLOWED
  • 31. ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 3. In how many ways can 3 runners finished the race if there are no ties? 2. In how many ways are there to place four different colored tiles in a row? 4. In how many ways can 7 different books can be arranged on a shelf if there is space enough for 4 books? 5. In how many ways can a president and vice president be chosen from a club with 10 members? 1. In how many ways are there to arrange the letters ABCD if repetition is allowed?
  • 32. ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 1. In how many ways are there to arrange the letters ABCD if repetition is allowed? 4 letters (All are taken and repetition is allowed) Given: 𝒏𝑷𝒓 = 𝒏𝒓 4𝑃4 = πŸ’πŸ’ 4𝑃4 = 4 βˆ™ 4 βˆ™ 4 βˆ™ 4 πŸ’π‘·πŸ’ = πŸπŸ“πŸ” π’˜π’‚π’šπ’” 𝑷(πŸ’, πŸ’) = πŸπŸ“πŸ” π’˜π’‚π’šπ’”
  • 33. ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 2. In how many ways are there to place four different colored tiles in a row? 4 colored tiles (All are taken and repetition is not allowed) Given: 𝒏𝑷𝒏 = 𝒏! 4𝑃4 = 4! 4𝑃4 = 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 πŸ’π‘·πŸ’ = πŸπŸ’ π’˜π’‚π’šπ’” 𝑷(πŸ’, πŸ’) = πŸπŸ’ π’˜π’‚π’šπ’”
  • 34. ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 3. In how many ways can 3 runners finished the race if there are no ties? 3 runners (All are taken and repetition is not allowed) Given: 𝒏𝑷𝒏 = 𝒏! 3𝑃3 = 3! 3𝑃3 = 3 βˆ™ 2 βˆ™ 1 πŸ‘π‘·πŸ‘ = πŸ” π’˜π’‚π’šπ’” 𝑷(πŸ‘, πŸ‘) = πŸ” π’˜π’‚π’šπ’”
  • 35. ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 4. In how many ways can 7 different books can be arranged on a shelf if there is space enough for 4 books? 7 different books (4 are taken and repetition is not allowed) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 7𝑃4 = 7! (7 βˆ’ 4)! 7𝑃4 = 7! 3! 7𝑃4 = 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3! 3! 7𝑃4 = 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 πŸ•π‘·πŸ’ = πŸ–πŸ’πŸŽ π’˜π’‚π’šπ’” 𝑷(πŸ•, πŸ’) = πŸ–πŸ’πŸŽ π’˜π’‚π’šπ’”
  • 36. ACTIVITY 1.10: LET’S TRY IT TOGETHER Answer the following questions. 5. In how many ways can a president and vice president be chosen from a club with 6 members? 6 members (2 are taken and repetition is not allowed) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 6𝑃2 = 6! (6 βˆ’ 2)! 6𝑃2 = 6! 4! 6𝑃2 = 6 βˆ™ 5 βˆ™ 4! 4! 6𝑃2 = 6 βˆ™ 5 πŸ”π‘·πŸ = πŸ‘πŸŽ π’˜π’‚π’šπ’” 𝑷(πŸ”, 𝟐) = πŸ‘πŸŽ π’˜π’‚π’šπ’”
  • 37. EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 2. In how many ways can the digits 1, 2, 3, 4, 5, and 6 be arranged to make a 3-digit code if repetition is allowed? 1. In how many ways can you arrange 8 figurines on a shelf? 3. In how many ways can 3 bicycles be parked if there are 5 available parking spaces? 4. In how many ways can Aling Maria arrange her 6 different potted plants in her plant stand if only 3 plants can fit?
  • 38. EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 1. In how many ways can you arrange 8 figurines on a shelf? 8 figurines (All are taken and repetition is not allowed) Given: 𝒏𝑷𝒏 = 𝒏! 8𝑃8 = 8! 8𝑃8 = 8 βˆ™ 7 βˆ™ 6 βˆ™ 5 βˆ™ 4 βˆ™ 3 βˆ™ 2 βˆ™ 1 πŸ–π‘·πŸ– = πŸ’πŸŽ, πŸ‘πŸπŸŽ π’˜π’‚π’šπ’” 𝑷 πŸ’, πŸ’ = πŸ’πŸŽ, πŸ‘πŸπŸŽ π’˜π’‚π’šπ’”
  • 39. EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 2. In how many ways can the digits 1, 2, 3, 4, 5, and 6 be arranged to make a 3-digit code if repetition is allowed? 6 digits (3 are taken and repetition is allowed) Given: 𝒏𝑷𝒓 = 𝒏𝒓 6𝑃3 = 63 6𝑃3 = 6 βˆ™ 6 βˆ™ 6 πŸ”π‘·πŸ‘ = πŸπŸπŸ” π’˜π’‚π’šπ’” 𝑷(πŸ”, πŸ‘) = πŸπŸπŸ” π’˜π’‚π’šπ’”
  • 40. EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 3. In how many ways can 3 bicycles be parked if there are 5 available parking spaces? 5 parking spaces (3 are taken and repetition is not allowed)) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 5𝑃3 = 5! (5 βˆ’ 3)! 5𝑃3 = 5! 2! 5𝑃3 = 5 βˆ™ 4 βˆ™ 3 βˆ™ 2! 2! 5𝑃3 = 5 βˆ™ 4 βˆ™ 3 πŸ“π‘·πŸ‘ = πŸ”πŸŽ π’˜π’‚π’šπ’” 𝑷(πŸ“, πŸ‘) = πŸ”πŸŽ π’˜π’‚π’šπ’”
  • 41. EVALUATION 1.3: MISSION POSSIBLE Answer the following questions. 4. In how many ways can Aling Maria arrange her 6 different potted plants in her plant stand if only 3 plants can fit? 6 potted plants (3 are taken and repetition is not allowed)) Given: 𝒏𝑷𝒓 = 𝒏! (𝒏 βˆ’ 𝒓)! 6𝑃3 = 6! (6 βˆ’ 3)! 6𝑃3 = 6! 3! 6𝑃3 = 6 βˆ™ 5 βˆ™ 4 βˆ™ 3! 3! 6𝑃3 = 6 βˆ™ 5 βˆ™ 4 πŸ”π‘·πŸ‘ = 𝟏𝟐𝟎 π’˜π’‚π’šπ’” 𝑷(πŸ”, πŸ‘) = 𝟏𝟐𝟎 π’˜π’‚π’šπ’”