Permutations And Combinations Ques 39
- 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 Adv.)
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Answer:
Correct Answer: 39.(5)
Solution:
Formula:
- 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 $={ }^{6} C _2$
4 girls are to be selected out of $5={ }^{5} C _4$
Now, 2 groups of girls can be arranged in 2 ! ways
Also, the group of 4 girls and 5 boys is arranged in $4 ! \times 5$ ! ways .
Now, total number of ways $\quad={ }^{6} C _2 \times{ }^{5} C _4 \times 2 ! \times 4 ! \times 5$ ! [from Eqs. (i), (ii), (iii) and (iv)]
$\therefore \quad m={ }^{6} C _2 \times{ }^{5} C _4 \times 2 ! \times 4 ! \times 5 !$
and $n=5 ! \times 6$ !
$\Rightarrow \quad \frac{m}{n}=\frac{{ }^{6} C _2 \times{ }^{5} C _4 \times 2 ! \times 4 ! \times 5 !}{6 ! \times 5 !}=\frac{15 \times 5 \times 2 \times 4 !}{6 \times 5 \times 4 !}=5$
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