Let n be the number of ways in which 5 boys a | vedclass

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(A) For $n$: Treat the $5$ girls as a single block. We have $5$ boys and $1$ block of girls,totaling $6$ entities. These can be arranged in $6!$ ways.

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$5$

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(A) For $n$: Treat the $5$ girls as a single block. We have $5$ boys and $1$ block of girls,totaling $6$ entities. These can be arranged in $6!$ ways. The $5$ girls can be arranged among themselves in $5!$ ways. Thus,$n = 6! 	imes 5!$.For $m$: We want exactly $4$ girls to stand consecutively. First,choose $4$ girls out of $5$ in $^5C_4 = 5$ ways.Treat these $4$ girls as a single block. We now have $5$ boys and $1$ block of $4$ girls,and $1$ remaining girl,totaling $7$ entities.To ensure exactly $4$ girls are together,we arrange these $7$ entities in $7!$ ways,then subtract cases where the $5$th girl is adjacent to the block of $4$ (which would make $5$ girls together). The number of ways to arrange $7$ entities such that the $5$th girl is not adjacent to the block is $(7! - 2 	imes 6!)$.Finally,multiply by the internal arrangements of the $4$ girls $(4!)$ and the $5$ ways to choose them: $m = 5 	imes (7! - 2 	imes 6!) 	imes 4! = 5 	imes (7 	imes 6! - 2 	imes 6!) 	imes 4! = 5 	imes 5 	imes 6! 	imes 4!$.Calculating $\frac{m}{n} = \frac{25 	imes 6! 	imes 4!}{6! 	imes 5!} = \frac{25 	imes 24}{120} = \frac{600}{120} = 5$.

Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue such that all the girls stand consecutively. Let $m$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue such that exactly four girls stand consecutively. Then the value of $\frac{m}{n}$ is

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