20 persons are invited for a party. In how ma | vedclass

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(B) Total number of persons $= 20 + 1 = 21$.Let the host be $H$ and the two particular persons be $P_1$ and $P_2$.Since $P_1$ and $P_2$ must sit on ei

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$2 	imes 18!$

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(B) Total number of persons $= 20 + 1 = 21$.Let the host be $H$ and the two particular persons be $P_1$ and $P_2$.Since $P_1$ and $P_2$ must sit on either side of the host,we treat the group $(P_1, H, P_2)$ as a single unit.This leaves $21 - 3 = 18$ other persons plus the $1$ unit,making a total of $19$ entities to be arranged in a circle.The number of ways to arrange $n$ items in a circle is $(n-1)!$,so $19$ entities can be arranged in $(19-1)! = 18!$ ways.Within the unit $(P_1, H, P_2)$,the two persons $P_1$ and $P_2$ can be arranged in $2! = 2$ ways (i.e.,$P_1HP_2$ or $P_2HP_1$).Therefore,the total number of ways is $2 	imes 18!$.

$20$ persons are invited for a party. In how many different ways can they and the host be seated at a circular table,if two particular persons are to be seated on either side of the host?

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