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2013 AMC 10 A #24

Video by Art of Problem Solving.  Problem from the MAA American Mathematics Competitions. Created by Art of Problem Solving.

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Video transcript

- Now in this problem, we have Central High School playing against Northern High School in a backgammon match. Each team has three players, and each player plays two games against each of the players on the other team, so each player's gonna play six games. That means the match is gonna take place in six rounds, and during each round, three games are gonna be played simultaneously. So I got Central over here, Northern over here, and in each round they're gonna pair off. These three are gonna pair off, play, and then separate. And next round they come in, they'll pair off again, and so on. And each player's gonna 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? Well in order to play around with this problem a little bit I'm gonna name the players. We're gonna call the Central High School players, they're gonna be A, B, and C. And the Northern High School players, they're gonna be M, N, and O. So I'm gonna start off here. Well I'm gonna take a bit of a what I call a constructive counting approach. I'm just gonna try to put together a schedule and see what I run into as I try to put together a schedule. I'm gonna think about the two rounds in which A is playing M. There's gonna be two different rounds in which these two have to play each other. Now what's gonna happen with the other four players? I guess I have a few options here. B, I can have B play 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'll happen, that's basically, it's gonna 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's set, then C is just gonna play whoever's left over there. So this is the way I'm gonna organize my counting. It's a very important first step is to get organized. And here we have a nice organization. Right here we're gonna 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 gonna have to multiply by two in the end, and I'm gonna go ahead and write that down, times two. We're gonna count this up and then multiply by two, 'cause whatever we get for this case is gonna be the same as 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. And then continuing on, in the last pair of rounds, well there'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. You know, this is the same as this, this is the same as this, this is the same as this. So rounds, 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 these rounds. And then we just drop these right in. Well we've got six letters there. So that's six factorial, but then we have to divide by two factorial for each of the repeats 'cause that six factorial will count, you know, overcounts, 'cause the order is XX, flip it over, still XX, still have the same schedule. So six factorial is 720. Two times two times two is eight. Divide by eight, that gives us 90. So that's 90 for this case. Now when we jump back here, we see that times two sitting out there and we remember to double it 'cause the case where B plays O in 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. We'll look at what happens when A is playing N. Well, while A is playing N, well what's gonna 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, 'cause B's already played O once. C can't play O in both rounds 'cause C's already played O once. So that means B is gonna have to play O in one round and C is gonna have to play O in the other 'cause 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 there are six 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.