In a row,6 boys and 6 girls are arranged at r | vedclass

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(D) Total number of ways to arrange $12$ people ($6$ boys and $6$ girls) is $12!$.To find the number of ways where $6$ girls are together,we treat the

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$1/132$

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(D) Total number of ways to arrange $12$ people ($6$ boys and $6$ girls) is $12!$.To find the number of ways where $6$ girls are together,we treat the $6$ girls as a single unit. Now,we have $6$ boys and $1$ unit of girls,totaling $7$ entities.These $7$ entities can be arranged in $7!$ ways.The $6$ girls within their unit can be arranged in $6!$ ways.So,the number of favorable arrangements is $7! 	imes 6!$.The probability $P$ is given by $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{7! 	imes 6!}{12!}$.$P = \frac{7! 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1}{12 	imes 11 	imes 10 	imes 9 	imes 8 	imes 7!} = \frac{720}{95040} = \frac{1}{132}$.

In a row,$6$ boys and $6$ girls are arranged at random. Find the probability that all $6$ girls are together.

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