Question

Find the number of ways in which $5$ boys and $5$ girls be seated in a row so that no $2$ girls may sit together.

Answer 1

Number of boys$=5$

Number of girls$=5$

$5$ boys can be seated in a row in ${=}^5{P}_5=5!$ ways

$\times B_1\times B_1\times B_1\times B_1\times B_1\times$

$5$ girls can be arranged in $6$ gaps in ${}^6{P}_5=\frac{5!}{(6-5)!}=6!$

Hence, the number of ways in which no two girls sit together$=5!\times 6!$ ways$=86,400‬$ways.