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Current Student
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A group of 8 friends want to play doubles tennis. How many [#permalink]
 02 Oct 2005, 02:51
Question Stats:
40% (02:04) correct
60% (02:23) wrong based on 0 sessions
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
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Current Student
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My answer was B, but we are both wrong. Any other takers?
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Senior Manager
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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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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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Current Student
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Close, but no cigar. Anybody else wish to take a crack??
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Manager
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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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I got 8C2*6C2*4C2*2C2 = 2520
so B is my pick.
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Current Student
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OA is E. Jdtomatito nailed it! 
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VP
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jdtomatito's explanation makes sense but I still don't understand why 8C2*6C2*4C2*2C2 is incorrect... 
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7C1 * 5C1 * 3C1 * 1C1
7*5*3*1 = 105
Answer is E?
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VP
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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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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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Intern
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sadsack 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.....
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sadsack 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)
Thanks for explanation. I feel better now 
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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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Intern
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sadsack wrote: 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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Director
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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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tingle wrote: sadsack 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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Current Student
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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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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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close
