A Group consists of 4 couples in which each of the 4 persons have one wife each. In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions?

Case I :  MW  MW  MW  MW

Case II:  WM  WM  WM  WM

Let us arrange 4 men in 4! ways, then we arrange 4 women in 4P4 ways at 4 places either left of the men or right of the men. Hence required number of arrangements

    $4! \times 4P4 + 4! \times 4P4 = 2 \times 576 = 1152$