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CIRCULAR
PERMUTATION
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
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,
๐‘ƒ = ๐‘› โˆ’ 1 !
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?
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)
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)
Thus, the permutation of n objects arranged in a
circle is
๐‘ƒ = (๐‘› โˆ’ 1)!
Wherein:
๐‘› โˆ’ ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘›๐‘œ. ๐‘œ๐‘“ ๐‘œ๐‘๐‘—๐‘’๐‘๐‘ก๐‘ 
To solve the given problem lets apply the formula
Given:
๐‘› = 3 (๐ฝ๐‘œ๐‘ ๐‘’, ๐‘ƒ๐‘Ž๐‘œ๐‘™๐‘œ, ๐‘Š๐‘Ž๐‘™๐‘™๐‘ฆ)
Solution:
๐‘ƒ = ๐‘› โˆ’ 1 !
๐‘ƒ = 3 โˆ’ 1 !
๐‘ƒ = 2!
๐‘ƒ = 2 ร— 1
๐‘ƒ = 2
Therefore, there are 2 possible ways we can
arrange Jose, Wally and Paolo if they will seat
in a round table.
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
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
๐‘ƒ๐‘› = ๐‘› โˆ’ 1!
Example 1:
Suppose 7 students are sitting around a
circle. Calculate the number of
permutations if clockwise and
anticlockwise arrangements are
different.
Solution:
๐‘ƒ๐‘›= ๐‘› โˆ’ 1!
๐‘ƒ๐‘›= 7 โˆ’ 1!
๐‘ƒ๐‘› = 6!
๐‘ƒ๐‘›= 6 โ‹… 5 โ‹… 4 โ‹… 3 โ‹… 2 โ‹… 1
๐‘ƒ๐‘›= 720
Hence, there are 720 possible arrangements of 7
students around a circle, given the fact that clockwise
and anticlockwise arrangements are different.
Example 2:
In how many ways can 8 people be
seated at a round table?
Solution:
๐‘ƒ๐‘›= ๐‘› โˆ’ 1!
๐‘ƒ๐‘›= 8 โˆ’ 1!
๐‘ƒ๐‘›= 7!
๐‘ƒ๐‘›= 7 โ‹… 6 โ‹… 5 โ‹… 4 โ‹… 3 โ‹… 2 โ‹… 1
๐‘ƒ๐‘›= 5040
Hence, 5040 different combinations are possible of 8
balls in a circle, given the fact that the clockwise and
anticlockwise arrangements are different.
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.
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
๐‘ƒ =
๐‘›โˆ’1!
2
(without lock/clasp) and
But if bracelets, key rings, and the like have a
lock, then the permutation becomes linear and
can be denoted as,
๐‘ƒ =
๐‘›!
2
(with lock)
where:
n - represents the number of objects in a set
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?
Solution:
a.
๐‘ƒ =
5โˆ’1!
2
๐‘ƒ =
4!
2
๐‘ƒ =
24
2
๐‘ƒ = 12 ๐‘ค๐‘Ž๐‘ฆ๐‘ 
Solution:
b.
๐‘ƒ =
5!
2
๐‘ƒ =
120
2
๐‘ƒ = 60 ๐‘ค๐‘Ž๐‘ฆ๐‘ 
Example 2:
Suppose 7 students are sitting around a
circle. Calculate the number of permutations
if clockwise and anticlockwise arrangements
are the same.
Solution:
๐‘ƒ =
7 โˆ’ 1!
2
๐‘ƒ =
6!
2
=
720
2
= 360
Hence, there are 360 permutations if 7
students are sitting around a circle given
that the clockwise and anticlockwise
arrangements are the same.
QUIZ TIME!
Direction: Write it on the 1
whole sheet of paper.
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.
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?
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.
THANK YOU!

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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, ๐‘ƒ = ๐‘› โˆ’ 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 ๐‘ƒ = (๐‘› โˆ’ 1)! Wherein: ๐‘› โˆ’ ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘›๐‘œ. ๐‘œ๐‘“ ๐‘œ๐‘๐‘—๐‘’๐‘๐‘ก๐‘  To solve the given problem lets apply the formula Given: ๐‘› = 3 (๐ฝ๐‘œ๐‘ ๐‘’, ๐‘ƒ๐‘Ž๐‘œ๐‘™๐‘œ, ๐‘Š๐‘Ž๐‘™๐‘™๐‘ฆ)
  • 8. Solution: ๐‘ƒ = ๐‘› โˆ’ 1 ! ๐‘ƒ = 3 โˆ’ 1 ! ๐‘ƒ = 2! ๐‘ƒ = 2 ร— 1 ๐‘ƒ = 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 ๐‘ƒ๐‘› = ๐‘› โˆ’ 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: ๐‘ƒ๐‘›= ๐‘› โˆ’ 1! ๐‘ƒ๐‘›= 7 โˆ’ 1! ๐‘ƒ๐‘› = 6! ๐‘ƒ๐‘›= 6 โ‹… 5 โ‹… 4 โ‹… 3 โ‹… 2 โ‹… 1 ๐‘ƒ๐‘›= 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: ๐‘ƒ๐‘›= ๐‘› โˆ’ 1! ๐‘ƒ๐‘›= 8 โˆ’ 1! ๐‘ƒ๐‘›= 7! ๐‘ƒ๐‘›= 7 โ‹… 6 โ‹… 5 โ‹… 4 โ‹… 3 โ‹… 2 โ‹… 1 ๐‘ƒ๐‘›= 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 ๐‘ƒ = ๐‘›โˆ’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, ๐‘ƒ = ๐‘›! 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. ๐‘ƒ = 5โˆ’1! 2 ๐‘ƒ = 4! 2 ๐‘ƒ = 24 2 ๐‘ƒ = 12 ๐‘ค๐‘Ž๐‘ฆ๐‘ 
  • 20. Solution: b. ๐‘ƒ = 5! 2 ๐‘ƒ = 120 2 ๐‘ƒ = 60 ๐‘ค๐‘Ž๐‘ฆ๐‘ 
  • 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: ๐‘ƒ = 7 โˆ’ 1! 2 ๐‘ƒ = 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.