Question

In how many ways can $6$ boys and $5$ girls can be seated in a row such that no two girls are together?

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

Answer 1

The correct option is C $6!{\times }^7{P}_5$

Since there is restriction on girls, let us seat the boys first, which can be done in $6!$ ways.

$\times B_1\times B_2\times B_3\times B_4\times B_5\times B_6\times$

For $5$ girls, there are $7$ places shown by ${}^{\prime \prime }{\times }^{\prime \prime }$ above $($note that here $2$ or more boys can sit together$)$ in which they can be seated in ${}^7{P}_5$ ways.

$\therefore$ Total number of ways ${=}^7{P}_5\times 6!$