QJEEAdv15Mad198: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Mad198 Solution: Let n be the number of ways in which 5 boys and 5 girls can…

Given: 5 boys and 5 girls

$n=$ number of ways of arranging them in a queue such that all the girls stand consecutively.

Let us consider 5 girls as one set

So, we have to arrange 5 boys and one set of girls. They can be arranged in 6 ! ways.

Also, the girls in the set can be arranged in 5 ! ways

So, total number of ways $=6 ! \times 5!$

$\Rightarrow n=6 ! \times 5 !$

Now, $m=$ number of ways of arranging them in a queue, such that exactly four girls stand consecutively.

$\because$ Exactly four girls can stand together so the remaining one girl must not stand consecutively with four girls.

Let us consider 2 cases:

Case I : The set of four girls is at the corner. Firstly, four girls are selected out of five girls in ${ }^5 \mathrm{C}_4$ ways. These girls are arranged in 4 ! ways.

Also, these girls can be placed in any of the two corners and the remaining one girl cannot stand next to the set of girls placed at the corner. So, the $5^{\text {th }}$ girl can stand at (7-1-1 = 5 ways) And the boys can be arranged in 5 ! ways.

$\begin{aligned}\text {So, number of ways } & =4 ! \times 2 \times{ }^5 \mathrm{C}_4 \times 5 ! \times 5 \\\\& =2 \times 5 \times 5 ! \times 5 !\end{aligned}$

Case II: The set of four girls are not placed at the corner.

So, four girls can be selected and arranged among themselves in ${ }^5 C_4 \times 4 !=5$ ! ways These girls are not at the corner so they can be arranged at 5 places.

The $5^{\text {th }}$ girl can stand at $7-2-1=4$ ways. $\{$ As she cannot stand at places near the set of four girls $\}$ and the boys can be arranged in 5 ! ways.

So, number of ways $=5 ! \times 5 \times 5 \times 4 \times 5!$

$\begin{aligned}\Rightarrow m & =(2 \times 5 \times 5 ! \times 5 !)+(5 \times 4 \times 5 ! \times 5 !) \\\\& =5 ! \times 5 !(10+20) \\\\& =30 \times 5 ! \times 5 ! \\\\\therefore \frac{m}{n} & =\frac{30 \times 5 ! \times 5 !}{6 ! \times 5 !} \\\\\Rightarrow \frac{m}{n} & \left.=\frac{30 \times 5 !}{6 \times 5 !} \quad \quad \because n !=n(n-1) !\right. \\\\\Rightarrow \frac{m}{n} & =5\end{aligned}$