# 2013 AMC 10 A #24

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Now in this problem, we have Central High School playing against Northern High School in a backgammon match. Each team has 3 players. And each player plays 2 games against each of the players on the other team. So each player is going to play 6 games. That means the match is going to take place in 6 rounds. And during each round, 3 games are going to be played simultaneously. So I got Central over here, Northern over here, and in each round, they're going to pair off. These two are going to pair off, play, and then separate. In the next round, they come in, they'll pair off again, and so on. And each player's going to play each player on the other team twice. We want to figure out how many different ways can the match be scheduled? How many different ways can we set up these six rounds? In order to play around with this problem a little bit, I'm going to name the players. We're going to call the Central High School players. They're going to be A, B, and C. And the Northern High School players, they're going to be M, N, and O. So I'm going to start off here, I'm going to take a bit of what I call a constructive counting approach. I'm just going to try to put together a schedule and see what I run into as I try to put together a schedule. I'm going to think about the two rounds in which A is playing M, two different rounds in which these two have to play each other. Now, what's going to happen with the other four players? Well, I guess I have a few options here. With B, I can have B playing N in both rounds. So B plays the same player. And it's basically the same thing as B playing O in both rounds. So if I can count how many ways this will happen, it's going to be the same thing as counting how many ways that'll happen. Or B can play different players in each of those two rounds. So these are my two options for B. And once B is set, then C is just going to play whoever's left over there. So this is the way I'm going to organize my accounting. It's a very important first step is to get organized. And here we have a nice organization. Right here we're going to look at the cases where B is playing the same player in both of these rounds and where B is playing different players. Now we have to remember back here, this case right here, we're going to have to multiply by 2 in the end. I'm going to go ahead and write that down, times 2. We're going to count this up and then multiply it by 2. Because whatever we get for this case, it's going to be the same for what we would get for that case. So let's focus on this first. We'll go ahead and look at that. We've got A is playing M. And B will play the same opponent in both rounds. And that leaves C with O. Now let's look at what happens when A is playing N in these two rounds. Well, who is B playing? Well, B could play O or M. That's not such a big deal. But C, well, C can't play O anymore. C's already played O twice, so C has to play M, which means B has to play O. And that's what they'll have to do in both of these. Continuing on, in the last pair of rounds, well, there's, it's obvious what each player has to do. They just have to play the remaining opponents they have left. So there are our six rounds. And as we can see, we've got the six rounds here, but they're in identical pairs. This is the same as this. This is the same as this. This is the same as this. So these two rounds are x's. These two rounds are y's. These two rounds are z's. So really, the only decision we have to make once we've set up all of these is just what order these come in. So what we're really doing is just ordering the word, xxyyzz. How many ways can we order this word? And that'll give us the order of the rounds. Then we just drop these right in. Well, we got six letters there, so that's 6 factorial. But then we have to divide by 2 factorial for each of the repeats. So that 6 factorial over counts because the orders xx, flip it over, still xx, still have the same schedule. So 6 factorial is 720. 2 times 2 times 2 is 8. Divide by 8, that gives us 90. So that's 90 for this case. And when we jump back here, we see that times 2 sitting out there. And we remember to double it because the case where B plays O on both rounds will also give us 90. That's 180 total. And now we move on to this other case here where B plays two different people while A is playing M. So we've got A versus M in these two rounds. B plays N in one of them. B plays O in the other. And then we know who C is playing. So we'll do the same thing we did before. We'll keep just trying to construct the rounds. Well, look at what happens when A is playing N. When A is playing N, well, what's going to happen here? Well, nobody can play N. Somebody has to play M. Somebody has to play O. Well, B can't play O in both rounds because B's already played O once. C can't play O in both rounds because C has already played O once. So that means B is going to have to play O in one round and C is going to have to play O in the other because O can't play either one of them twice. And that fills out the schedule. And we just keep on going just like this. Look at what happens when A is playing O. Well, B has to play M and N. That's all that's left. Plays M in one round, N in the other. Can only play one game with each. And then we know what C is doing in each of these rounds. And these six rounds, they're all different. So now we know that these are the six rounds that have to be played when A is playing M, B is playing different players, while A is playing M. These are the six rounds that have to occur. All that matters is the order of these six rounds. They're all different, so their 6 factorial equals 720 ways to order these six rounds. So we go back here. Our second case down here, there are 720 possibilities. We add these two together, we get 900, and we're done.