Discussion Forum

 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

Answer:

Explanation:

Here,   $-B_{1}-B_{2}-B_{3}-B_{4}-B_{5}-$

  out of 5 girls , 4 girls are together and 1 girl is seperate  . 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! X 5!  ways .....(iv)

Now, total number of ways  $=^{6}C_{2}\times ^{5}C_{4}\times 2!\times4!\times5!$

        [ from Eqs. (i),(ii) ,(iii) and (iv)]

$\therefore$       m$=^{6}C_{2}\times ^{5}C_{4}\times 2!\times4!\times5!$

  and                 n= 5! X6!

 $\Rightarrow$  $\frac{m}{n}=\frac{^{6}C_{2}\times ^{5}C_{4}\times 2!\times4!\times5!}{6!\times5!}$

                       = $\frac{15\times5\times2\times4!}{6\times5\times4!}=5$