The total number of ways in which 4 boys and 4 girls can form a line, 

with boys and girls alternating, is

The four girls can first be arranged in $4!$ am,1ng the rnseives. In each of these arrangements there are :,::aps (including the extremes) c1mong the girls. Since the bov and girls are lo alternate, we have to leave the first gap or Igap
blank while arranging the boys. But, in each case the bov

and girls Gm b'-' arranged in $4!\cdot 4!$ ways.

$\therefore$ Required number of ways  $=2(4!\times 4!)=2(4!{)}^2.$