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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.

## Want to join the conversation?

- How do you know when to use ! ?(10 votes)
- Well when you are trying to calculate the amount of combinations of something. Let's say you have 1,2, and 3. You have 3 slots to place it in to make a 3 digit number. For the first slot we have 3 possible. For the second only 2 because we used 1 up, and then the last slot you only have one choice. The total amount of possibilities is 3 X 2 X 1 which is 3!. This is called combinatorics.(1 vote)

- What if A is playing different people per round

eg.

Instead of A-M, A-M in one round

isn't A-M, A-O, or some other combo like that possible?(8 votes)- When the instructor does the last step in each tree, he counts the different ways he can reorder the "word" made up by his made combinations: "xxyyzz" can be reordered 90 different ways (see3:40to4:20). This accounts for all those other combos.(5 votes)

- how do you do roots(2 votes)
- it is the opposite of a square, normally people just use a calculator(5 votes)

- What does vactoreol mean! = - 0(2 votes)
- its factorial and it basically means that n! = n x (n-1) x (n-2) ... 1. All the way to one. For example 3! = 3 x 2 x 1(1 vote)

- at4:00, what method does he uses to find the combination?(2 votes)
- I know this is a little bit late, but here it is:

http://www.artofproblemsolving.com/videos/counting/chapter3/59

Richard explains it really well. I recommend Aops videos and books if you can handle it.(3 votes)

- what did he mean at5:34?(3 votes)
- At that point, Person A still needs to play two games with Person O, so these must comprise the top "level" of the round (although it should be noted that the "level" does not matter, as the three games are simultaneous).

Both Persons B and C need to play one game each with Persons M and N. Since one person can not be in two games at the same time, the choices are B-M/C-N or B-N/C-M.

These two rounds (of three games each) do not have to occur last, which is why he uses 6! to find how many ways the six rounds can form the match. :)(2 votes)

- At4:10how did he get what he did, i don't understand(2 votes)
- Why did he divide 720 by 2!2!2! at3:59? And why are factorials involved?(2 votes)
- Is this some sort of test on KA?

If it is, when does it happen? It seems kind of interesting to me and I would be grateful to participate in this event.(1 vote)- What Viha said is pretty good advice. You could also Google either test you want to take and there will likely be information about it somewhere, even on Wikipedia. If there is a math related club at your school, ask your moderator about joining. My high school had a Mu Alpha Theta society and so I did AMC10 with them. Anyways, here's how I understand the AMC tests when I took it...there are three different tests for three different proficiency levels: AMC8, AMC10, and AMC12. These are named because the AMC8 test is built to challenge eighth graders, AMC10 for tenth graders, and AMC12 for twelfth graders. These tests consist of 25 multiple choice questions with difficulties ranging from child's play to #REKT. You probably won't need to study any topics other than what you have covered in your years studying math. If you're in high school, I highly suggest taking the test. I took it with no cost, and our moderator told us that we could receive monetary rewards for exceptional scores. I did not receive any awards, even though I got second best score out of the ten who took the test from our high school...anyways, happy math-ing, comrade!(2 votes)

- do they still do these? i once saw they had a math contest, or something like that... how do you get to those?(1 vote)

## 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.