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 fou

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 $\frac{m}{n}$ is [2015]

Smart Achievers · Accepted Answer

(5)

$Here,_B_{1}_B_{2}_B_{3}_B_{4}_B_{5}$

Out of 5 girls, 4 girls are together and 1 girl is separate. Now, to select 2 positions out of 6 positions between boys $= {}^{6}C_{2} ...(i)$

4 girls are to be selected out of 5 $= {}^{5}C_{4} ...(ii)$

Now, 2 groups of girls can be arranged in $2!$ ways $. ..(iii)$

Also, the group of 4 girls and 5 boys is arranged in $4! \times 5!$ ways $...(iv)$

Now, total number of ways $= {}^{6}C_{2} \times {}^{5}C_{4} \times 2! \times 4! \times 5!$ $[from Eqs. (i), (ii), (iii) and (iv)]$

$∴$ $m = {}^{6}C_{2} \times {}^{5}C_{4} \times 2! \times 4! \times 5!$ $and n = 5! \times 6!$

$\Rightarrow \frac{m}{n} = \frac{{}^{6}C_{2} \times {}^{5}C_{4} \times 2! \times 4! \times 5!}{6! \times 5!} = \frac{15 \times 5 \times 2 \times 4!}{6 \times 5 \times 4!} = 5$

Solution

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