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Question

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


Solution

Let us calculate $$n$$ first. 

We can consider $$5$$ girls as one set and the $$5$$ boys. We can arrange them in $$6!$$ ways. The girls in the set can be arranged in $$5!$$ ways. 
Hence, $$n= 5! \times 6! $$
Let us calculate $$m$$. 

Now, let us first place the 4 girls together. We will need to consider different cases. 

Case 1 : $$4$$ girls at the corner. $$4$$ girls can be selected and permuted in $$^5P_4 = 5 !$$ ways. The $$5^{th}$$ girl can be placed in $$5$$ ways. The boys can be placed in $$5!$$ ways. They can be placed in the left or the right corner. 

Cases $$ = 2 \times 5 \times 5! \times 5 ! $$
Case 2: The $$4$$ girls are not placed at the corner. 
The position of the $$4$$ girls together can be selected in $$5$$ ways. The $$4$$ girls can be selected and permuted in $$^5P_4 $$ ways. The $$5^{th}$$ girl can be placed in $$4$$ ways. The $$5$$ boys can be placed in $$5!$$ ways. 

Cases $$= 5\times 5! \times 4 \times 5 ! $$

Hence, $$ m = 30 \times 5 ! \times 5 ! $$
Hence, $$\dfrac{m}{n} = 5 $$

Maths

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