Permutations - Problem 4

We're now going to look at two different ways to organize data and they're very, very similar in theory, but the answers are going to be very, very different. So the first one we're going to look at is we're trying to arrange four guys and four girls in such a way that guys and girls have to alternate.

So when you're dealing with guys and girls each guy is going to be different, I don't have them named up here because they'd be a lot more writing, but basically we have four different guys, four different girls and we can rearrange our guys and it will be a different order completely.

So what we can think about is 2, 4, 6, 7, 8 so what we picked is one gender has to go first it doesn't matter who we'll say a guy. So we have four guys to choose from, any of these four guys can go first which tells us we have four people who can go here.

The next spot is a girl and we have four different girls, so any of those four will be able to go second, so we have guy, girl, our next one has to be a guy. One guy is already in place, so we only have three left to choose from, so that tells us we have three remaining. Same thing for the girls, somewhere down the road we've already used two guys already to get to this point so then we have two guys that can go here, two girls can go here one guys, one girl.

So what we end up in this case is 4 factorial times 4 factorial, but one other thing we have to consider is that here I assumed that guys go first, that doesn't have to be the case, there is no stipulation that said guys have to go first, so what we could do is have a girl first in which case the number of options is still going to be exactly the same, four girls could go first then four guys could go second and so on and so forth.

So what basically we could do is say this times 2 to include the guys going first and the girls going first, so what we end up having we could Foil this out or multiply this out, I'm only concerned with the number I'm concerned with the concept so what we end up here is two groups alternating four factorial times four factorial times 2. A very similar question, but extremely different answer is going to be the case with balls, so in this case I chose to write it blue, colors don't really matter, but we're trying to arrange four red balls and four blue balls so that they are alternating as well.

So for this one 5, 6, 7, 8 we can choose a color to go first either red or blue let's say red and the trick for this one is that these are supposed to be balls that are not distinguishable, so the red is the red you're not going to be able to tell the difference at all and so basically one red is here, one red is here we're not going to tell the difference so there is really one way that this will look if red goes first because they're all going to look the same.

There is another way that you can look if blue goes first, so here we could just say red, red, red, red, and then the blues fill in the gap we're not going to be able to tell the difference. Similarly we could do it the other way where we could say blue goes first and then we have red going second, fourth, sixth and last. Again we're not going to be able to tell the difference from those because these balls look exactly the same.

So in this case what we're really left with is just 2, so this is the difference between a non-distinguishable situation and a distinguishable situation. With the kids we can tell them apart. Two kids don't necessarily look the same so therefore we can rearrange them a little different, while we're dealing with the ball situation, we rearrange those where you can't really tell the difference so therefore we only have two situations that we're dealing with.