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

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

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$2 	imes 18!$

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(B) Total number of persons = $20$ (guests) $+ 1$ (host) $= 21$ persons.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 $H$,we can treat $(P_1, H, P_2)$ as a single block.Within this block,the arrangement can be $(P_1, H, P_2)$ or $(P_2, H, P_1)$,which is $2! = 2$ ways.After treating the three as one unit,the remaining number of persons is $21 - 3 = 18$. Including the one block,we have $18 + 1 = 19$ units to arrange in a circle.The number of ways to arrange $n$ items in a circle is $(n-1)!$. Here,$n = 19$,so the number of ways is $(19-1)! = 18!$.Total ways $= 2 	imes 18!$.

$20$ 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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