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### CIRCULAR-PERMUTATION-Copy.pptx

• 2. At the end of the lesson, the students should be able to: a. define circular permutation; b.solve problems involving circular permutation; c. relate circular permutations in real-life situations. OBJECTIVES
• 3. CIRCULAR PERMUTATION Circular permutation is the arrangement of objects in a circular manner. It is the total number of ways in which n distinct object can be arranged around a fixed circle defined as, 𝑃 = 𝑛 − 1 !
• 4. Analyze this: Suppose it happens that (1) Jose, (2) Wally and (3) Paolo will visit you in your house, how can you arrange them in a round table if you will prepare them a snack?
• 5. These are 3 possible arrangements if we will arrange them in clockwise position: (1) Jose, (2) Wally, (3) Paolo (2) Wally, (3) Paolo, (1) Jose (3) Paolo, (1) Jose, (2) Wally 1 3 2 (a)
• 6. These are the 3 possible arrangements if we will arrange them in counter clockwise position: (1) Jose, (3) Paolo, (2) Wally (3)Paolo, (2) Wally, (1) Jose (2) Wally, (1) Jose, (3) Paolo 1 3 2 (b)
• 7. Thus, the permutation of n objects arranged in a circle is 𝑃 = (𝑛 − 1)! Wherein: 𝑛 − 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑜. 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 To solve the given problem lets apply the formula Given: 𝑛 = 3 (𝐽𝑜𝑠𝑒, 𝑃𝑎𝑜𝑙𝑜, 𝑊𝑎𝑙𝑙𝑦)
• 8. Solution: 𝑃 = 𝑛 − 1 ! 𝑃 = 3 − 1 ! 𝑃 = 2! 𝑃 = 2 × 1 𝑃 = 2 Therefore, there are 2 possible ways we can arrange Jose, Wally and Paolo if they will seat in a round table.
• 9. a. When clockwise and counter-clockwise orders are different There are two types of circular permutation: b. When clockwise and counter-clockwise orders are the same
• 10. a. When clockwise and counter-clockwise orders are different/ If the clockwise and counter-clockwise orders CAN be distinguished, then the total number of circular permutations of n elements taken all together is 𝑃𝑛 = 𝑛 − 1!
• 11. Example 1: Suppose 7 students are sitting around a circle. Calculate the number of permutations if clockwise and anticlockwise arrangements are different.
• 12. Solution: 𝑃𝑛= 𝑛 − 1! 𝑃𝑛= 7 − 1! 𝑃𝑛 = 6! 𝑃𝑛= 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 𝑃𝑛= 720 Hence, there are 720 possible arrangements of 7 students around a circle, given the fact that clockwise and anticlockwise arrangements are different.
• 13. Example 2: In how many ways can 8 people be seated at a round table?
• 14. Solution: 𝑃𝑛= 𝑛 − 1! 𝑃𝑛= 8 − 1! 𝑃𝑛= 7! 𝑃𝑛= 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 𝑃𝑛= 5040 Hence, 5040 different combinations are possible of 8 balls in a circle, given the fact that the clockwise and anticlockwise arrangements are different.
• 15. b. Observe the arrangement of different beads in a bracelet, keys on the key rings, and the like. The clockwise and the counter-clockwise orders are not distinguishable.
• 16. So, when clockwise and counter-clockwise orders are the same/ If the clockwise and counter-clockwise orders CANNOT be distinguished, then the total number of circular permutations of n elements taken all together is 𝑃 = 𝑛−1! 2 (without lock/clasp) and
• 17. But if bracelets, key rings, and the like have a lock, then the permutation becomes linear and can be denoted as, 𝑃 = 𝑛! 2 (with lock) where: n - represents the number of objects in a set
• 18. Example 1: In how many ways can 5 different beads be arranged if: a. a bracelet has no lock? b. a bracelet has a lock?
• 19. Solution: a. 𝑃 = 5−1! 2 𝑃 = 4! 2 𝑃 = 24 2 𝑃 = 12 𝑤𝑎𝑦𝑠
• 21. Example 2: Suppose 7 students are sitting around a circle. Calculate the number of permutations if clockwise and anticlockwise arrangements are the same.
• 22. Solution: 𝑃 = 7 − 1! 2 𝑃 = 6! 2 = 720 2 = 360
• 23. Hence, there are 360 permutations if 7 students are sitting around a circle given that the clockwise and anticlockwise arrangements are the same.
• 25. Direction: Write it on the 1 whole sheet of paper.
• 26. 1. In how many ways can 10 keys be arranged on a key ring if a key ring has no lock? 2. 11 boy scouts are scattered around a camp fire. How many ways can they be arranged? 3. In how many ways can eight different beads be arranged on a bracelet if a bracelet has a lock? 4. Find the number of different ways that a family of 7 can be seated around a circular table with 7 chairs. 5. 9 students are sitting around a circle. Calculate the number of permutations if clockwise and anticlockwise arrangements are the same.
• 27. 1. How many ways can 10 different colored toy horses be arranged on a merry-go-round? 2. In how many ways can 6 different beads be arranged if: a. a bracelet has no lock? b. a bracelet has a lock?
• 28. In your notebook, try to answer the following questions: 1. President Ferdinand Marcos Jr. together with his 4 Cabinet members is holding a press conference with 6 media reporters. How many different ways can they be seated in a round table if the president and his cabinet members must sit next to each other? 2. Give one example of problems or situations in real life that involve circular permutations. In your example, a. explain the problem or situation. b. solve the problem.
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