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.We arrange $6$ boys in $6!$ ways. This creates $7$ gaps (including ends) where $5$ girls can be placed.Numbe

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$3$

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(C) Case $(I)$: No two girls sit together.We arrange $6$ boys in $6!$ ways. This creates $7$ gaps (including ends) where $5$ girls can be placed.Number of ways $p = 6! 	imes {}^{7}P_{5} = 6! 	imes \frac{7!}{2!} = 6! 	imes \frac{7 	imes 6 	imes 5 	imes 4 	imes 3}{1} = 6! 	imes 2520$.Case $(II)$: All girls sit together.Treat the $5$ girls as a single unit. Now we have $6$ boys and $1$ unit of girls,total $7$ entities.These $7$ entities can be arranged in $7!$ ways. The $5$ girls within the unit can be arranged in $5!$ ways.Number of ways $q = 7! 	imes 5!$.Ratio $p/q = \frac{6! 	imes \frac{7!}{2!}}{7! 	imes 5!} = \frac{6!}{2! 	imes 5!} = \frac{720}{2 	imes 120} = \frac{720}{240} = 3$.

$6$ 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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