  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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