Question

in how many ways can 4 boys and 3 girls stand in a row so that no two girls are together?

Answer 1

$<b><i>Hey MATE!$

Let us first take 4 places and make the boys stand in those 4 places by 4! ways.

Now to put girls so that no 2 girls are together, we put girls in 5 gaps between 5 boys.

This is done by
$\binom{4}{3} \times 3 \: factorial$
So the total ways of doing the whole work is :

=> 5! * (4 C 3) * 3!

=> 576 Ways

Hope it helps

Hakuna Matata :))