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

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

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02 Oct 2005, 02:51
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Question Stats:

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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?

A. 420
B. 2520
C. 168
D. 90
E. 105

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html
[Reveal] Spoiler: OA

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02 Oct 2005, 04:35
My answer was B, but we are both wrong. Any other takers?

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02 Oct 2005, 07:04
8C2*6C2*4C2*2C2 = 2520 ways 8 people will form 4 teams of 2 each
My guess is we need to divide this number by 4C2 (two teams from 4 possible teams can play a game of doubles) ...
Hence, 2520/4C2 = 2520/6 = 420

Thats the only explanation I can think of... I must admit initially I thought the answer was 2520 too...
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02 Oct 2005, 20:34
Antmavel wrote:
C 168 for me

2C8*2C4*2C2 = 28*6*1 = 168

Stupid Of course my answer is wrong : I forgot 2C6
Anyway, I would have found 2520 like you...

Sofinally I am still lost if B is not the OA

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02 Oct 2005, 23:18
Close, but no cigar. Anybody else wish to take a crack??

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03 Oct 2005, 02:01
I think the answer is E 105.

If you pick one player, he has 7 possible partners.
We have six left, if we pick one, he has 5 possible partners.
We have four, if we pick one, he has 3 different possible partners,

Therefore 7*5*3=105

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03 Oct 2005, 02:13
I got 8C2*6C2*4C2*2C2 = 2520

so B is my pick.
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hey ya......

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03 Oct 2005, 07:07
OA is E. Jdtomatito nailed it!

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03 Oct 2005, 08:11
jdtomatito's explanation makes sense but I still don't understand why 8C2*6C2*4C2*2C2 is incorrect...

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03 Oct 2005, 12:08
7C1 * 5C1 * 3C1 * 1C1
7*5*3*1 = 105

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03 Oct 2005, 13:12
jdtomatito wrote:
I think the answer is E 105.

If you pick one player, he has 7 possible partners.
We have six left, if we pick one, he has 5 possible partners.
We have four, if we pick one, he has 3 different possible partners,

Therefore 7*5*3=105

Its difficult for me to understand.

If you pick one player, he has 7 possible partners.

here, first player could be picked from pool of 8 people each having 7 remaining people as partner. All combinations seems 8*7 rather than just 7. Dont know what I am missing.

I would selected B.

8c2(first team) * 6c2(second team) * 4c2(third team) * 2c2 (last team)

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03 Oct 2005, 14:16
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This is way I solved it :

( (8c2) * (6c2) * (4c2) * (2c2) ) / 4! = 105

The 4! is needed because the order of the 4 teams is unimportant. For example the following pairs are equivalent...

(1,2) (3,4) (5,6) (7,8) <==> (3,4) (1,2) (7,8) (5,6)

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03 Oct 2005, 15:45
This is way I solved it :

( (8c2) * (6c2) * (4c2) * (2c2) ) / 4! = 105

The 4! is needed because the order of the 4 teams is unimportant. For example the following pairs are equivalent...

(1,2) (3,4) (5,6) (7,8) <==> (3,4) (1,2) (7,8) (5,6)

Order isnt imp and isnt that the reason why we have used combination instead of Permutation here??? why do we need to divide the outcome by 4! pls explain.....

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03 Oct 2005, 16:14
This is way I solved it :

( (8c2) * (6c2) * (4c2) * (2c2) ) / 4! = 105

The 4! is needed because the order of the 4 teams is unimportant. For example the following pairs are equivalent...

(1,2) (3,4) (5,6) (7,8) <==> (3,4) (1,2) (7,8) (5,6)

Thanks for explanation. I feel better now

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03 Oct 2005, 16:16
There's two things going on here:
(1) ordering between members in a team.
(2) ordering between teams

In this question order is NOT relevant on both counts. Using combinations ensures ordering between members is excluded.
Dividing by 4! makes sure ordering between teams is also excluded.
I hope this helps

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03 Oct 2005, 17:28
There's two things going on here:
(1) ordering between members in a team.
(2) ordering between teams

In this question order is NOT relevant on both counts. Using combinations ensures ordering between members is excluded.
Dividing by 4! makes sure ordering between teams is also excluded.
I hope this helps

thanks sadsack for clearing the confusion... its a great help indeed!

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03 Oct 2005, 18:40
With that one, it's painfully obvious that I'm totally lost with permutations and combinations. Anybody got any suggestions? I've gone through the GMAT club course material, and a couple of other books too. Still my mind just refuses to think in the proper way!!!

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03 Oct 2005, 19:58
tingle wrote:
This is way I solved it :

( (8c2) * (6c2) * (4c2) * (2c2) ) / 4! = 105

The 4! is needed because the order of the 4 teams is unimportant. For example the following pairs are equivalent...

(1,2) (3,4) (5,6) (7,8) <==> (3,4) (1,2) (7,8) (5,6)

Order isnt imp and isnt that the reason why we have used combination instead of Permutation here??? why do we need to divide the outcome by 4! pls explain.....

I had the same problem Tingle. This was my issue. thanks sadsack for the clear post.

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03 Oct 2005, 22:59
anandsebastin> I too am completely dumbfounded when it comes to these thypes of problems, which is exactly why I keep posting them over and over again, scrutinizing the methodology. Have you checked out the book Veritas Project GMAT yet? I have heard that they really drill in this subject matter.

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02 Jan 2006, 00:50
In other words, how many different ways can be used to pick 2 players out of 8. Of course, the order does not matter.

C(8;2)=28
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02 Jan 2006, 00:50

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