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boksana
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There are 8 people who want to divide themselves to play 04 Jul 2004, 14:26
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There are 8 people who want to divide themselves to play double tennis into 4 teams. How many combinations are possible?



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There are 8 people who want to divide themselves to play 04 Jul 2004, 14:35
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i don't think so :roll:
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Paul
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There are 8 people who want to divide themselves to play 04 Jul 2004, 14:35
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First team: 8C2
Second team: 6C2
Third team: 4C2
Fourth team: 2C2
total outcome: 8C2*6C2*4C2*2C2 = 2520
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boksana
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There are 8 people who want to divide themselves to play 04 Jul 2004, 14:36
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Nope Paul! Try to solve for 4 people and you'll see your mistake
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boksana
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There are 8 people who want to divide themselves to play 04 Jul 2004, 14:50
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Are you kidding? :)
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Paul
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There are 8 people who want to divide themselves to play 04 Jul 2004, 15:04
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:stupid I think you have to divide by 4! because each team could be chosen 1st, 2nd, 3rd or 4th irrespective of order. Hence 2520/4! = 105
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There are 8 people who want to divide themselves to play 04 Jul 2004, 15:07
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:stupid2 :shocked :hammer
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gmatblast
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There are 8 people who want to divide themselves to play 10 Aug 2004, 11:07
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Would it be appropriate to say that if the question had been for the number of different possible teams, then Paul's answer would have been correct?

Here the question asks for how many different combinations? I believe that if the 4 teams are formed as (1,2)(3,4),(5,6) and (7,8), then this whole configuration will constitute one combination.

But as for the number of teams, it is 4 teams

Am I correct?



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