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(D) Total number of ways to arrange $8$ teachers and $4$ students around a circular table is $(8+4-1)! = 11!$. To ensure no two students sit together,

Vedclass · Accepted Answer

$\frac{7}{33}$

Vedclass · Answer

(D) Total number of ways to arrange $8$ teachers and $4$ students around a circular table is $(8+4-1)! = 11!$. To ensure no two students sit together,we first arrange the $8$ teachers in a circle,which can be done in $(8-1)! = 7!$ ways. This creates $8$ gaps between the teachers. We need to place $4$ students in these $8$ gaps,which can be done in $^8C_4$ ways. The students can be arranged among themselves in $4!$ ways. Thus,the number of favorable arrangements is $^8C_4 	imes 4! 	imes 7!$. The required probability is $\frac{^8C_4 	imes 4! 	imes 7!}{11!} = \frac{\frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} 	imes 4! 	imes 7!}{11 	imes 10 	imes 9 	imes 8 	imes 7!} = \frac{70 	imes 24 	imes 7!}{11 	imes 10 	imes 9 	imes 8 	imes 7!} = \frac{7}{33}$.

$8$ teachers and $4$ students are sitting around a circular table at random. Then,the probability that no two students sit together is:

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