See Fig.
1st part: Total persons at the circular table = 20 guests + 1 host = 21 They can be seated in (21 − 1)!, that is, 20! ways.
2nd part: After fixing the places of three persons (1 host + 2 persons), treating (1 host + 2 person) as 1 unit, we have now 19 {(remaining 18 persons + 1 unit) = 19} and the number of arrangement will be (19 − 1)! = 18!. Also, these two particular persons can be seated on either side of the host in 2! ways.
Hence, the number of ways of seating 21 persons at the circular table such that two particular persons be seated on either side of the host is 18! x 2! = 2 x 18! ways.
The circular permutations in which clockwise and anticlockwise arrangements give rise to same permutations, for example, arranging some beads to form a necklace.
Figure. 1.
Consider five beads A, B, C, D and E in a necklace or five flowers A, B, C, D and E in a garland, etc. If the necklace or the garland on the left is turned over we obtain the arrangement on the right, that is, anticlockwise and clockwise order of arrangements are not different. We will get arrangements as follows:
We can see that the arrangements are not different.
Then the number of circular permutations of n different things taken all at a time is 1/2(n -1)! , if clockwise and anticlockwise orders are not taken as different.
