Combinatorics task. 
Four men and four women shall get in line in a supermarket. In how many ways can they line up, if the line has to alternate between men and women (two men and women can not stand next to each other)?

My solution: $4 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1 = 576$; Then we multiplicate it by two because a man or a woman can be first in line; $576 \cdot 2 = 1152$.
The answer given in my maths book: $2304$.
What did I do wrong?

You may as well separate them into two lines and call them up one at a time, alternating between the lines. You can arrange each line in $4!$ ways. You can alternate between the lines starting from either the male or female line, so that it $2$ choices. Meaning there are $(4!)^2 \cdot 2= 1152$ total possibilities, agreeing with your answer. It is probably a typo, the solutions having doubled twice.
