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Permutations & Combinations

Permutations And Combinations

MODULES

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Basics of Permutations
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Basics of Combinations
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Letters Technique
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Using Blanks
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Blanks with Repetition
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Slotting Technique
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Conditional Permutations
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Special Types
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Circular Permutations
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Derangement
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Binomial Expansion
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Forming Groups
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Identical Elements
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Identical Groups
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Selection in Geometry
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Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Permutations & Combinations 1
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Permutations & Combinations 2
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Permutations & Combinations 3
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Permutations & Combinations 4
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Permutations & Combinations 5
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Permutations & Combinations 6
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PRACTICE

Permutations & Combinations : Level 1
Permutations & Combinations : Level 2
Permutations & Combinations : Level 3
ALL MODULES

CAT 2025 Lesson : Permutations & Combinations - Circular Permutations

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4. Circular Permutations

When elements are arranged in a circle, there is no starting point or ending point. So, to convert the circular arrangement into a linear arrangement, we take one of the elements, fix it to its place and compute the number of ways the rest of the elements can be permuted.

If n distinct elements are arranged in a circle, the number of such distinct arrangements is (n−1)!\bm{(n - 1)!}(n−1)!

If this circle can be reversed, number of such arrangements is (n−1)!2\bm{\dfrac{(n - 1)!}{2}}2(n−1)!​ . This is because we do not need to count those arrangements that are produced when a circle is flipped.

Example 33

In a group of 666 people, two of them have a quarrel. In how many ways can these 666 people be seated around a circular table such that the two who are quarrelling are never together?

(1) 484848                      (2) 727272                      (3) 120120120                      (4)144144144                     

Solution

666 people can be seated around a circular table in (6−1)!=5!=120(6 - 1)! = 5! = 120(6−1)!=5!=120 ways

We can consider the two quarrelling people as one entity, which leaves us with 555 people to be seated, wherein the 222 quarrelling people can be permuted in 2!2!2! ways within their group.
When 222 people are always together =(5−1)!×2!=48= (5 - 1)! \times 2! = 48=(5−1)!×2!=48 ways

∴\therefore∴ Number of ways the 222 are never seated together =120−48=72= 120 - 48 = \bm{72}=120−48=72

Alternatively (Slotting)

The 444 people who are not in quarrel can be seated in (4–1)!(4 – 1)!(4–1)! ways

Now, there are 444 slots between them to seat the two people. The 222 slots can be selected in 4C2^{4}C_{2}4C2​ ways and the two quarrelling people can be arranged in these slots in 2!2!2! ways.

∴\therefore∴ Number of ways the 222 are never seated together =(4−1)!×4C2×2!=6×6×2=72= (4 - 1)! \times ^{4}C_{2} \times 2! = 6 \times 6 \times 2 = \bm{72}=(4−1)!×4C2​×2!=6×6×2=72

Answer: (222) 727272


Example 34

There is a meeting of five Indian and five British diplomats. The 101010 diplomats are to be seated around a circular table such that each Indian diplomat has a British diplomat on either side. How many ways can the 101010 diplomats be seated?

(1) 5!×4!5! \times 4!5!×4!                      (2) 4!×4!4! \times 4!4!×4!                      (3) 10!10!10!                      (4) 9!9!9!                     

Solution

We can start by seating the 555 Indian diplomats around the circular table. This can be done in (5−1)!=4!(5 - 1)! = 4!(5−1)!=4! ways.

Now, there are 555 slots between the Indians, where the 555 British diplomats can be seated. We can select the 555 seats in 5C5^{5}C_{5}5C5​ ways and arrange the British diplomats in 5!5!5! ways.

∴\therefore∴ Number of ways the 101010 diplomats can be seated =4!×5C5×5!=5!×4!= 4! \times ^{5}C_{5} \times 5! = \bm{5! \times 4!}=4!×5C5​×5!=5!×4!

Answer: (111) 5!×4!5! \times 4!5!×4!

Note 1) There is only 111 way to select 555 from 555 available slots, i.e. 5C5=1^{5}C_{5} = 15C5​=1. Going forward, if nnn elements are to be selected from nnn available elements, you can ignore these in your calculations as there is only 111 way to do this.

2) A common mistake in these questions is to choose (5−1)!×(5−1)!=4!×4!(5 - 1)! \times (5 - 1)! = 4! \times 4!(5−1)!×(5−1)!=4!×4! . Note that in any circular permutation question, once we fix an element, the circle is broken. Therefore, (n−1)!(n - 1)!(n−1)! is applied only once.


Example 35

555 Indian and 222 British diplomats are to sit around a table, in such a way that the 222 British diplomats do not sit next to each other. In how many different ways can they be seated?

(1) 720720720                      (2) 600600600                      (3) 480480480                      (4) 360360360                     

Solution

Number of ways in which 555 Indian diplomats can be placed =(5−1)!=24= (5 - 1)! = 24=(5−1)!=24 ways.

We select 222 of the 555 slots and arrange the 222 British officers in 5C2×2!=2^{5}C_{2} \times 2! = 25C2​×2!=2 ways

∴\therefore∴ Total Permutations =24×20=480= 24 \times 20 = 480=24×20=480

Answer: (3) 480480480


Example 36

Josie needs to insert 444 different gems into an empty string and wear it for a function. In how many ways can the gems be arranged to form different necklace patterns?

(1) 3 33                      (2) 444                      (3) 666                      (4) 121212                     

Solution

The 444 different gems can be arranged in (4−1)!(4 - 1)!(4−1)! ways. But when the necklace is inverted and seen, it would look the same. So, each arrangement produces another one when flipped.

∴\therefore∴ Different arrangements can be produced by 444 gems in (4−1)!2=62=3\dfrac{(4 - 1)!}{2} = \dfrac{6}{2} = \bm{3}2(4−1)!​=26​=3 ways

Answer: (1) 333


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