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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)!}

If this circle can be reversed, number of such arrangements is (n−1)!2\bm{\dfrac{(n - 1)!}{2}} . 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 66 people, two of them have a quarrel. In how many ways can these 66 people be seated around a circular table such that the two who are quarrelling are never together?

(1) 4848                      (2) 7272                      (3) 120120                      (4)144144                     

Solution

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

We can consider the two quarrelling people as one entity, which leaves us with 55 people to be seated, wherein the 22 quarrelling people can be permuted in 2!2! ways within their group.
When 22 people are always together =(5−1)!×2!=48= (5 - 1)! \times 2! = 48 ways

∴\therefore Number of ways the 22 are never seated together =120−48=72= 120 - 48 = \bm{72}

Alternatively (Slotting)

The 44 people who are not in quarrel can be seated in (4–1)!(4 – 1)! ways

Now, there are 44 slots between them to seat the two people. The 22 slots can be selected in 4C2^{4}C_{2} ways and the two quarrelling people can be arranged in these slots in 2!2! ways.

∴\therefore Number of ways the 22 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}

Answer: (22) 7272


Example 34

There is a meeting of five Indian and five British diplomats. The 1010 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 1010 diplomats be seated?

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

Solution

We can start by seating the 55 Indian diplomats around the circular table. This can be done in (5−1)!=4!(5 - 1)! = 4! ways.

Now, there are 55 slots between the Indians, where the 55 British diplomats can be seated. We can select the 55 seats in 5C5^{5}C_{5} ways and arrange the British diplomats in 5!5! ways.

∴\therefore Number of ways the 1010 diplomats can be seated =4!×5C5×5!=5!×4!= 4! \times ^{5}C_{5} \times 5! = \bm{5! \times 4!}

Answer: (11) 5!×4!5! \times 4!

Note 1) There is only 11 way to select 55 from 55 available slots, i.e. 5C5=1^{5}C_{5} = 1. Going forward, if nn elements are to be selected from nn available elements, you can ignore these in your calculations as there is only 11 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! . Note that in any circular permutation question, once we fix an element, the circle is broken. Therefore, (n−1)!(n - 1)! is applied only once.


Example 35

55 Indian and 22 British diplomats are to sit around a table, in such a way that the 22 British diplomats do not sit next to each other. In how many different ways can they be seated?

(1) 720720                      (2) 600600                      (3) 480480                      (4) 360360                     

Solution

Number of ways in which 55 Indian diplomats can be placed =(5−1)!=24= (5 - 1)! = 24 ways.

We select 22 of the 55 slots and arrange the 22 British officers in 5C2×2!=2^{5}C_{2} \times 2! = 2 ways

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

Answer: (3) 480480


Example 36

Josie needs to insert 44 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 3                      (2) 44                      (3) 66                      (4) 1212                     

Solution

The 44 different gems can be arranged in (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 44 gems in (4−1)!2=62=3\dfrac{(4 - 1)!}{2} = \dfrac{6}{2} = \bm{3} ways

Answer: (1) 33


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