Question

$20$ persons are invited to a party.Let "k" the number of ways where the persons invited and the host be seated at a circular table. Let "m" be the number of ways where two particular persons be seated on either side of the host. Find $\frac{k}{10m}$ ?

Answer 1

$Istpart:\text{Total persons on the circular table}=20guests+1host=21$
They can be seated in $(21-1)!=20!ways$
$IIndpart:$ After fixing the places of three persons (1 host + 2 persons). Treating these 3 as a unit, we now have (remaining 18 people and 1 unit = 19). And the number of arrangement will be $(19-1)!=18!$ also these two particular persons can be seated 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 = $18!\times 2!=2\times 18!$