In how many arrangements can 7 men and 7women be placed in a line if the women and men alternate.
I thought the would 14! divided by something but not finding it...
$\displaystyle \boxed{MW}\;\boxed{MW}\;\ldots \;\boxed{MW}\rightarrow{7!\cdot}7!$
$\displaystyle \boxed{WM}\;\boxed{WM}\;\ldots \;\boxed{WM}\rightarrow{7!\cdot}7!$
...
Fernando Revilla
He used "\boxed" in LaTex. You can see the LaTex code for any expression by double clicking on it.
One way to get that answer is to think of putting the 7 men in a row- there are 7! ways to do that- then put the 7 women in a row- there are 7! ways to do that. Then place the women, in that order, between each of the men in the row. Since any of the 7! ways of ordering the men can be matched with any of the 7! ways of ordering the women, there are (7!)(7!) ways to do that.
Another way to see that is to first pair each man with a woman- because of the "MW" boxes I suspect this is what FernandoRevilla did. There are 7 ways to pair a woman with the first man, 6 ways with the second, etc. so 7! pairs. There are then 7! ways to order those pairs.
However, that is always with a man first. If, instead, we put a woman first, then a man, etc., we have a second way for each such combination. I think the correct answer is (2)(7!)(7!).
I realize now that I misread the previous posts and that (2)(7!)(7!) was given all along.
Right, $\displaystyle (7!7!)2 =\ldots=?$
For example, if you write \boxed{\int_a^bf(x)\;dx} it will appear:btw how did you put a box accent around your text?
$\displaystyle \boxed{\int_a^bf(x)\;dx}$
Fernando Revilla
Edited: Sorry, I didn't see the previous post.