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# A group of 8 friends want to play doubles tennis. How many

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Senior Manager
Joined: 07 Oct 2003
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A group of 8 friends want to play doubles tennis. How many [#permalink]

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28 Aug 2004, 20:38
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A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

here's my flawed stab at a problem:
8C2*6C2*4C2 = 840

I'm getting a wrong answer. can someone point out the flaw in my reasoning?
Manager
Joined: 11 Jul 2004
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29 Aug 2004, 10:22
you are forgetting to divide the whole by 4! since the teams can be made up in 4! ways and we really dont care who is in which team

let me know if that gets you to the corect ans
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Joined: 07 Jul 2004
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29 Aug 2004, 22:58
Since order is not important, we just want to group them into pairs. So the asnwer should just be 8C2 = 28
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29 Aug 2004, 22:59
forgot to add... if you pair them up, you'll get 4 teams since there are only 8 people.
Intern
Joined: 03 Aug 2004
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30 Aug 2004, 11:13
for the uneducated, can you please tell me what the C stand for in 8C2...thanks
CIO
Joined: 09 Mar 2003
Posts: 463

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30 Aug 2004, 11:31
That's a very good question. I'm always worried that people who don't know what that means will get turned off. The truth is that it's just some jargon to denote the mathematical process of combinations. It's all quite easy, actually, if you just get away from the math and get into the concept. I've got it all written down here in a pretty concise way. But in order to understand combinations, you have to start with permutations.

What permutations are is this: when you've got to figure out the number of arrangements of a set of objects, for example, the number of ways 5 people can sit in 5 chairs, you have a permutations question. You always solve them the same way: lay out the number of spots to fill up, put the number of objects that can go into each spot, and then multiply. The answer to that chair question is 5x4x3x2x1, because for each chair, you have one less object to go, since someone is already sitting.

You could also have the same question when you've got more objects than spaces. For example, how many ways can 8 people sit in 3 chairs? 8x7x6.

But then you've got to ask yourself if the order matters. In the chair question, the order does matter because we want to know all the ways they can sit together, and perhpas it matters that the first guy sits in the first chair, etc. So we count all the permutations, and permutations are specific to when the order matters.

Sometimes the order doesn't matter. If I had 8 pizza toppings and I was going to make a three topping pizza, the answer would still by 8x7x6, but in this case I have to recognize that onions, mushrooms, and pepperoni is the same as pepperoni, onions, and mushrooms. So I don't want to count all the orders, because the order doens't matter. I just want to know how many pizzas I can make.

The short answer to that (and all you need to know, really) is that you'll take the permutations and divide by the (number of spaces) factorial. So in this case, because there are 3 spaces, the correct answer is (8x7x6)/3! = (8x7x6)/(3x2x1) = 56.

And that's what this is all about. The notation for combinations (the appropriate word when order doesn't matter) is to use C (it stands for "choose"). So in the pizza example, it would simply be enough to say 8C3, meaning, 8 objects, choosing 3, order doesn't matter. Then, anyone reading that would know that it's really (8x7x6)/3!.

Does that make sense?
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
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GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

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31 Aug 2004, 05:03
I do not understand this "order matters or not" thing I am sorry.
Same problem with Kaplan material as well

if you assume that n>p and nAp = n!/(n-p)! and nCp = n!/(p!(n-p)!), which is the basic maths I remember from school

nAp is always > nCp

orders matters should drop the number of combinations and be represented by nCp when nAp represents the non matters case.

In this tennis team dividing, I would have chosen 8A2 = 7*8=56 but it seems that it is wrong !
CIO
Joined: 09 Mar 2003
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31 Aug 2004, 09:08
Well, I think you're confused about two different things, so let me try to clear it up.

First, permutations will always be larger than combinations. It's sometimes counter intuitive. I've had many students who think that if the order matters, there should be less arrangements than if it doesn't matter. But it's definately the other way around.

When the order matters, that means we want to count ALL THE ORDERINGS. So there will be more of them. Think about a committee of 3 people with a president, vice president, and treasurer. If we're going to choose three people from a group of 8 people, there would be 336 committees to create. That's permutations - the order matters because person A would like to be president and not vp. So if he, and person B and person C are on the committee, then the order ABC is different than BCA, and so we have to count them both, and every other ordering of A, B, and C (there are 6 of them).

If we don't care about the order, if, let's say, the committee is based on equal voting, then we only care that ABC is a different committee than BCD, but it's not different than CBA. So we want to get rid of all the copies and just count the number of different groups. In that case, the answer of 336 would be divided by 3!, since that's the number of spaces there are, and the answer would be 56.

So that's the difference between Permutations and Combinations.

The reason that the answer to the question here isn't 8x7 is twofold. First, within each team, we are not differentiating between positions. 2 people on a team, and we don't care who's the first person and who's the second person, so we have to divide by 2. If the question had been written differently, if it had said that there was a designated server, then we'd have used permutations, because the order would have mattered. But not in this case. Here it's combinations.

But that's not enough, because we're not just choosing 2 people for a single team out of 8 people, we're choosing 4 teams of 2 out of 8 people. So once we've got the first team figure out (28 ways) there are still 6 people left for the other three teams. So we do it again for the second team - 6x5/2 - and the third team - 4x3/2 - and the 4th team - 2x1/2, and then we multiply all these together: 28x15x6x1.

Finally, we have to see all the teams as a separate combinations problem. Once we've got them, we don't care which teams goes first in the lineup, for example, nor are we putting them on the board in any given way. So we take the answer, and divide it by 4!, since there are 4x as many arrangements as there are combinations. We get four, because there are 4 teams.

Intern
Joined: 31 Aug 2004
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31 Aug 2004, 19:16
I cudn't understand y v r supposed to divide by 4!......8 pppl......4 grps of 2.......i.e 8C2*6C2*4C2........y shud v divide by 4!.......plzzz clarify!!!!!
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31 Aug 2004, 21:11
gmat800 wrote:
I cudn't understand y v r supposed to divide by 4!......8 pppl......4 grps of 2.......i.e 8C2*6C2*4C2........y shud v divide by 4!.......plzzz clarify!!!!!

This is because the teams of 2 are just set up with a random group. In this question, we are not assigning the teams of 2 to team A, B, C and D whereby the groups are distinct.
Think of it this way. If you have 4 flags, 2 red and 2 blue, how many distinct signs, raising them all four, can you do with those 4 flags? The answer would be 4! / 2!*2!. Here, we are dividing by 2!*2! because there are 2 groups of red/blue flags which are interchangeable among themselves. For instance, if you have the first and second positions being red, it does not matter which red flag you are putting up because either are red flags. Hence, wherever those red flags are, you can interchange them and the same applies for the blue flags.
Going back to the original problem, since the 4 teams can be interchanged because they are just a team and are in that respect not different, we have to divide by 4!.
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Best Regards,

Paul

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Joined: 31 Aug 2004
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31 Aug 2004, 23:17
thank u soooo much paul!! i get it now ...!
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

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31 Aug 2004, 23:51
I find it strange that by just naming the teams you need to do a division by 4!.

If you split 8 people into pairs, you get 4 teams, and even if you name the teams as team A, team B, team C, and team D, it should not make a difference. It's just a name afterall.

How you split the 8 people into the 4 teams (with or without name) should be the same number of ways.
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

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31 Aug 2004, 23:52
Paul, I"m sorry I didn't get the interchangeble flag explanation. Any easier way to understand?
CIO
Joined: 09 Mar 2003
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01 Sep 2004, 00:17
There seems to be a lot of confusion over this one. Let me try my hand at explaining it.

Everyone's on board with the first part, that we have 8c2x6c2x4c2x2c2 (using the shorthand for combinations), which equals 2520.

So we see that there are 2,520 teams ways these 8 people could be put in 8 teams.

Let's look closer at one possible variation. Let's say there's a red team, a blue team, an orange team, and a yellow team. One possible lineup of those four teams is: R, B, O, Y. That's 1 out of 2520, leaving 2519 to go. Now let's keep the same people in each team, but rearrange them B, Y, O, R. That's another possible lineup, leaving 2518 to go.

How many ways can we rearrange these four teams? 4!, that's how many.

So we need to ask the question. Does it matter how we line up these four teams? Are they in the finals and squaring off against each other? No. Are they ranked in the newspaper? No. Is there a rule that Red must play first? No. Simply, is there anything in the world that would differentiate between the following two orderings:

R, B, O, Y
B, Y, O, R

No. There's not.

So the order doesn't matter. And if we have one ordering, that's enough - we don't want to count all the others. So instead of looking at 2,520 different orderings, we want to divide that by 4!, so we get rid of all the copies.

That leaves 105.

Now, if you're versed in combinations, you may ask why there are any orderings at all? After all, when the number of objects equals the number of spaces, and it's a combinations question, there should just be one combination. 4C4 = 1, for example.

But remember that within the teams, the players are switching around. In one instance, persons 1 and 2 are on the blue team, and in another, persons 1 and 3 are in it. So the 105 orderings we've arrived at tells us exactly how many different ways these 8 people can go into these different teams, without copying them around at all. You can see the difference it makes on the final number.

I hope that helps. Let me know if there are areas still unclear.
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01 Sep 2004, 00:50
Ian, I think I get it now, we're removing the redundant extra team combinations since order is not important. However if order is critical, we do not divide it by 4!.

I like to digress a little here. Say order is important, such that now it matters that in Team Apple, John and Mary is not the same as Mary and John (since they have different order). So the number of ways to arrange 4 teams of 2 (with order) now becomes 8P2*6P4*4P2*2P2/4!. (Divide by 4! as the order of the team does not matter). Is that it ?
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Joined: 27 Jul 2004
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01 Sep 2004, 05:51

http://www.gmatclub.com/phpbb/viewtopic.php?t=9272
Director
Joined: 31 Aug 2004
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01 Sep 2004, 06:02
Dear Ian,

Thank you for your time and explainations. I've got it now!

Here is the copy of a post of Bhai whose url is reported above

1. In how many ways can 12 students be partitioned into 3 teams, A, B, and C, so that each team contains 4 students?

2. In how many ways can 12 students be divided into 3 teams of 4 people each?

If I correctly understood answer 2 is 12C4.8C4.4C4/3!
where answer 1 is 12C4.8C4.4C4 as groups are distinct

Do you agree ?
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01 Sep 2004, 06:43
twixt
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Best Regards,

Paul

01 Sep 2004, 06:43
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