Question

The number of ways in which $5$ Boys and $5$ Girls can be arranged in a row so that no two girls are together is

• A. $10!$
• B. $5!6!$
• C. $(5!{)}^2$
• D. $2(5!{)}^2$

Answer 1

The correct option is B $5!6!$

x $B_1$ x $B_2$ x $B_3$ x $B_4$ x $B_5$ x

$6\times$'s are there, for $5$ girls to sit.
So, the total no. of ways in which the boys can be seated $=5!$
So, the total no. of ways in which the girls can be seated $={}^6{P}_5=6!$

The total no. of ways $=6!\times 5!$