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 $5!\times 6!$. 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.
Best Answer
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?