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 $\frac{m}{n}$ ?___

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Solution

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,
$n o o f w a y s t o s e l e c t 2 p o s i t i o n s o u t o f 6 p o s i t i o n s b e t w e e n b o y s =^{6} C_{2} . . . (i)$
4 girls are to be selected out of 5 = $^{5} C_{4}$ ways …(ii)
Now, 2 groups of girls can be arranged in 2! ways. …(iii)
$A l s o, t h e g r o u p o f 4 g i r l s a n d 5 b o y s i s a r r a n g e d i n 4! \times 5! w a y s . . . (i v)$
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)]
$∴ m =^{6} C_{2} \times^{5} C_{4} \times 2! \times 4! \times 5! A n d n = 5! \times 6! \Rightarrow \frac{m}{n} = \frac{^{6} C_{2} \times^{5} C_{4} \times 2! \times 4! \times 5!}{6! \times 5!} = \frac{15 \times 5 \times 2 \times 4!}{6 \times 5 \times 4!} = 5$

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