If 3 sisters and 8 brothers are together play | vedclass

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(D) Total number of people = $3 + 8 = 11$. Number of ways to arrange $11$ people around a circle = $(11 - 1)! = 10!$. Now,consider the case where all

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$8! 	imes 84$

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(D) Total number of people = $3 + 8 = 11$. Number of ways to arrange $11$ people around a circle = $(11 - 1)! = 10!$. Now,consider the case where all $3$ sisters sit together. Treat the $3$ sisters as $1$ unit. Total units to arrange = $8$ brothers + $1$ unit of sisters = $9$ units. Number of ways to arrange $9$ units around a circle = $(9 - 1)! = 8!$. The $3$ sisters can be arranged among themselves in $3! = 6$ ways. So,the number of ways where all $3$ sisters sit together = $8! 	imes 6$. The number of ways where all $3$ sisters are not seated together = (Total arrangements) - (Arrangements where all $3$ sisters sit together) = $10! - (8! 	imes 6)$. $10! - 6 	imes 8! = (10 	imes 9 	imes 8!) - (6 	imes 8!) = (90 - 6) 	imes 8! = 84 	imes 8!$.

If $3$ sisters and $8$ brothers are together playing a game,then the number of ways in which all the sisters and brothers are to be seated around a circle such that all the $3$ sisters are not seated together is

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