Question

The number of ways in which $6$ Boys and $5$ Girls can sit in a row so that all the girls may be together is

• A. $6!\times 5!$
• B. $6{!}^7{P}_5$
• C. $(6!{)}^2$
• D. $7!\times 5!$

Answer 1

The correct option is D $7!\times 5!$

$5$ Girls $-1$ box.

So, there are total of $7$ boxes to rearrange.

$G_1{G}_2{G}_3{G}_4{G}_5$

$B_1$

$B_2$

$B_3$

$B_4$

$B_5$

$B_6$

No. of ways $=7!$
$\Rightarrow$ Also the girls can rearrange within the box by $5!$ ways.
$\Rightarrow$ Total no. of ways $=5!\times 7!$