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# 2013 AMC 10 A #24

Video transcript

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.