Arrange $n$ men and $n$ women in a row, alternating women and men. A group contains n men and n women. How many ways are there to arrange these people in a row if the men and women alternate?
I got as far as:
There are $n$ [MW] blocks. So there are $n!$ ways to arrange these blocks.
There are $n$ [WM] blocks. So there are $n!$ ways to arrange these blocks.
Which makes the total ways to arrange the men and women in alternating blocks $2(n!)$
The correct answer is $2(n!)^2$
Where does the second power come from?
 A: Imagine that you line up the men first; you can do this in $n!$ ways. Now you line up the women; you can do this in $n!$ ways as well. Finally, you mesh the two lines together so that the sexes alternate; you can do this in $2$ ways, depending on whether you put a man or a woman at the left end of the line. These are independent choices, so they can be combined in $2\cdot n!\cdot n!=2n!^2$ ways.
A: You added where you needed to multiply.  You're going to arrange $n$ men AND $n$ women in a row, not $n$ men OR $n$ women, so you've got $n!$ ways to do one task and $n!$ ways to do the other, making $(n!)^2$.
But after that there's this issue: Going from left to right, is the first person a man or a woman?  You can do it either way, so you have $(n!)^2 + (n!)^2$.
A: The number of ways to arrange the men in order is $n!$ Likewise for the number of women. So if you don't care whether a man or woman starts the row, the answer would $(n!)^2$, since you need to consider each possible permutation of men with any possible permutation of women. Since you do, multiply by 2.
A: Your solution doesn't take into account that $[M_1,W_1]$ and $[M_1,W_2]$ are valid combinations.
