6 Women and 5 Men number of positions problem I don't understand

I have my discrete math final coming up on monday and am trying to figure out how to do a few problems. The one I am having the most problem with is just very confusing because I don't know how to go about solving it, much less solving it. Here is the question

A group contains 5 men and 6 women. how many ways are there to arrange these people in a row if the men and women alternate? Hint: arrange the women first.

part of me thinks of doing a combination C(6,5) where there are 6 groups and 5 need to be ordered. However, a friend says to multiply 5!*6!. Then my teacher has another long explantation that I can't even follow because he runs through it so fast

I have no doubt this will be on the test, but I just don't understand how solve it.

-
Do you see why they suggest "arranging" the women first? (No, it's not a fashion question.) – Raskolnikov Apr 29 '11 at 19:35

Okay - because the men and the women must alternate, we know that the only possible configuration is for them to line up W M W M W M W M W M W, i.e. alternate with women on either end.

So now we can split this up into two smaller problems: how many ways can we arrange the 6 women, and how many ways can we arrange the 5 men? Well, a permutation on n objects can happen in $n!$ ways (there are n choices for the first object, n-1 for the second, and so on). So there are $6!$ ways of arranging the women and $5!$ ways of arranging the men. As there is no way for the WMW order to change, there are no other ways that the pattern can change.

Thus the answer is $6! * 5!$, as your friend said.

I also wanted to comment of your intuition of performing $C(6,5)$. That would men that you are only ordering 5 elements, but we know that we are lining up all 11! So I wonder how you were taught to think of combinations and permutations?

-
That makes a lot of sense and is very much what he said, but then he started second guessing himself. As for the combinations and permutations. That has been my eternal struggle for the last couple months. I can do the math formulaicly. Give me P(5,2) or C(20,3) I can do that. However, I still have a problem extrapolating from a world problem what goes where and how it fits together. At times it just doesn't make sense. – percent20 Apr 29 '11 at 19:43

I believe the answer given by mixedmath is wrong. Close, but wrong. I know this is old but it shows up on the first page of results when you google the question. He doesn't take into account that either a woman can start first or a man can start first, so I believe the correct answer is 2 * (6! * 5!).

EDIT: Sorry about that, yes he was right, and I was wrong. I was solving a similar problem that had an equal number of men and women and without thinking applied it to this one. My apologies.

-
A man cannot start first - there aren't enough men (try it). – Sarastro Dec 5 '12 at 6:02
This is no longer an answer. It should be deleted. – robjohn Dec 5 '12 at 9:43