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?
Best Answer
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 $(n!)^2 + (n!)^2$.