Question

The number of ways in which 5 boys and 3 girls can be seated in a row so that each girl in between two boys

• A. 2880
• B. 1880
• C. 3800
• D. 2800

Answer 1

The correct option is A 2880

First, we seat 5 boys, which can be done in $5!$ ways $B\times B\times B\times B\times B$. Three girls can be seated in places marked ${}^{\prime }{x}^{\prime }$ in ${}^4{P}_3$ways.
$\therefore$ Total ways $=5!{.}^4{P}_3=2880$