6 boys and 5 girls sit in a line such that (I | vedclass

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(C) Case $(I)$: No two girls sit together.First,arrange $6$ boys in a row,which can be done in $6!$ ways.There are $7$ gaps created between and at the

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$3$

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(C) Case $(I)$: No two girls sit together.First,arrange $6$ boys in a row,which can be done in $6!$ ways.There are $7$ gaps created between and at the ends of the boys $(	imes B_1 	imes B_2 	imes B_3 	imes B_4 	imes B_5 	imes B_6 	imes)$.We need to place $5$ girls in these $7$ gaps,which can be done in $^7P_5$ ways.So,$p = 6! 	imes ^7P_5 = 6! 	imes \frac{7!}{2!} = 6! 	imes \frac{7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2!}{2!} = 6! 	imes 2520$.Case $(II)$: All girls sit together.Treat the $5$ 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 $5$ girls within the unit can be arranged among themselves in $5!$ ways.So,$q = 7! 	imes 5!$.Calculating $p/q$:$\frac{p}{q} = \frac{6! 	imes ^7P_5}{7! 	imes 5!} = \frac{6! 	imes \frac{7!}{2!}}{7! 	imes 5!} = \frac{6!}{2! 	imes 5!} = \frac{720}{2 	imes 120} = \frac{720}{240} = 3$.

$6$ boys and $5$ girls sit in a line such that $(I)$ no two girls sit together $(II)$ all the girls sit together. If $p$ is the number of arrangements in case $(I)$ and $q$ is the number of arrangements in case $(II)$,then $p/q =$

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