If 6 girls and 5 boys are sitting in a row,wh | vedclass

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(C) Total number of ways to arrange $11$ people ( $6$ girls and $5$ boys) in a row is $11!$. To ensure no two boys sit together,we first arrange the $

Vedclass · Accepted Answer

$\frac{6! 	imes 7!}{2! 	imes 11!}$

Vedclass · Answer

(C) Total number of ways to arrange $11$ people ( $6$ girls and $5$ boys) in a row is $11!$. To ensure no two boys sit together,we first arrange the $6$ girls in $6!$ ways. This creates $7$ gaps (including ends) where the $5$ boys can be placed: $\_ G \_ G \_ G \_ G \_ G \_ G \_$. The number of ways to choose $5$ gaps out of $7$ and arrange the $5$ boys is $^7P_5 = \frac{7!}{2!}$. Thus,the number of favorable arrangements is $6! 	imes \frac{7!}{2!}$. The probability is $\frac{6! 	imes 7!}{2! 	imes 11!}$.

If $6$ girls and $5$ boys are sitting in a row,what is the probability that no two boys sit together?

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