A host invites 20 persons for a party. The number of ways they can be seated at a circular table such that two particular persons be seated on either side of the host is

Explanation

I part total persons on the circular table
=20 guest+1 host=21
They can be seated in (21-1)!=20! Ways ltbr. II part after fixing the places of three person
(1 host+2 person).
Treating (1 host+2 person)=1 unit, so we have now {(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 on the circular table such that two particular persons be seated on either side of the host=`18!xx2!=2xx18!`
Case II if clockwise and anti-clockwise orders are taken as not different, then the number of circular permutations of n different things taken r at a time `=(.^(n)P_(r))/(2r)=(1)/(2r)*(n!)/((n-r)!)`.