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)!
If this circle can be reversed, number of such arrangements is 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 6 people, two of them have a quarrel. In how many ways can these 6 people be seated around a circular table such that the two who are quarrelling are never together?
(1) 48
(2) 72
(3) 120
(4)144
Solution
6 people can be seated around a circular table in (6−1)!=5!=120 ways
We can consider the two quarrelling people as one entity, which leaves us with 5 people to be seated, wherein the 2 quarrelling people can be permuted in 2! ways within their group.
When 2 people are always together =(5−1)!×2!=48 ways
∴ Number of ways the 2 are never seated together =120−48=72
Alternatively (Slotting)
The 4 people who are not in quarrel can be seated in (4–1)! ways
Now, there are 4 slots between them to seat the two people. The 2 slots can be selected in 4C2 ways and the two quarrelling people can be arranged in these slots in 2! ways.
∴ Number of ways the 2 are never seated together =(4−1)!×4C2×2!=6×6×2=72
Answer: (2) 72
Example 34
There is a meeting of five Indian and five British diplomats. The 10 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 10 diplomats be seated?
(1) 5!×4!
(2) 4!×4!
(3) 10!
(4) 9!
Solution
We can start by seating the 5 Indian diplomats around the circular table. This can be done in (5−1)!=4! ways.
Now, there are 5 slots between the Indians, where the 5 British diplomats can be seated. We can select the 5 seats in 5C5 ways and arrange the British diplomats in 5! ways.
∴ Number of ways the 10 diplomats can be seated =4!×5C5×5!=5!×4!
Answer: (1) 5!×4!
Note
1) There is only 1 way to select 5 from 5 available slots, i.e. 5C5=1. Going forward, if n elements are to be selected from n available elements, you can ignore these in your calculations as there is only 1 way to do this.
2) A common mistake in these questions is to choose (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)! is applied only once.
Example 35
5 Indian and 2 British diplomats are to sit around a table, in such a way that the 2 British diplomats do not sit next to each other. In how many different ways can they be seated?
(1) 720
(2) 600
(3) 480
(4) 360
Solution
Number of ways in which 5 Indian diplomats can be placed =(5−1)!=24 ways.
We select 2 of the 5 slots and arrange the 2 British officers in 5C2×2!=2 ways
∴ Total Permutations =24×20=480
Answer: (3) 480
Example 36
Josie needs to insert 4 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
(2) 4
(3) 6
(4) 12
Solution
The 4 different gems can be arranged in (4−1)! ways. But when the necklace is inverted and seen, it would look the same. So, each arrangement produces another one when flipped.
∴ Different arrangements can be produced by 4 gems in 2(4−1)!=26=3ways