A group of 8 friends want to play doubles tennis. How many : PS Archive
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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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17 Jun 2008, 08:10
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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?

I read the C(8,2) * C(6,2) * C(4,2) * C(2,2)/4! approach

What if we modify the question in following way:

Total friends: 8
We need: 3 teams of 2, 2 and 4

What will the approach in this case?
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Re: PS: teams [#permalink]

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26 Jun 2008, 05:41
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First of all, which question are you referring to? There was the original question and then a subsequent question that was modified. The second question was 3 teams, First = 2 people, Second = 2 people and the 3rd = 4 people (total of 8) people.

Can you tell us which question you're talking about so we can explain ourselves?

rino wrote:
I dont see why shouldn't it be a simple 8c2.

i think u are all wrong

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings SVP Joined: 30 Apr 2008 Posts: 1887 Location: Oklahoma City Schools: Hard Knocks Followers: 40 Kudos [?]: 570 [0], given: 32 Re: PS: teams [#permalink] ### Show Tags 17 Jun 2008, 12:25 Quote: Total friends: 8 We need: 3 teams of 2, 2 and 4 What will the approach in this case? C(8,2) * C(6,2) * C(4,4) / 3! Why would it be incorrect? The first team of is a combination of 2 things taken from 8, or C(8,2), then the next is a team of 2 taken from 6 available choices, so C(6,2), and the final team you state is a team of 4 taken from 4 available choices C(4,4). We have a total of 3 teams and we do not want to treat the order of the teams differently so we divide that by 3!. Dividing by 3! removes the number of ways that each team can be ordered because unless we do that, we count the same teams, but in different order as a different combination, which is not what the question asks for. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: PS: teams [#permalink]

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17 Jun 2008, 18:48
jallenmorris wrote:

Why would it be incorrect?

The first team of is a combination of 2 things taken from 8, or C(8,2), then the next is a team of 2 taken from 6 available choices, so C(6,2), and the final team you state is a team of 4 taken from 4 available choices C(4,4). We have a total of 3 teams and we do not want to treat the order of the teams differently so we divide that by 3!. Dividing by 3! removes the number of ways that each team can be ordered because unless we do that, we count the same teams, but in different order as a different combination, which is not what the question asks for.

Good explanation Jallenmorris..
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Re: PS: teams [#permalink]

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18 Jun 2008, 06:25
C(8,2) * C(6,2) * C(4,4) / 3!

For the explanation to the answer see my post above.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Re: PS: teams [#permalink]

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18 Jun 2008, 20:07
jallenmorris wrote:
C(8,2) * C(6,2) * C(4,4) / 3!

For the explanation to the answer see my post above.

I see the explanation but no answer.
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Re: PS: teams [#permalink]

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26 Jun 2008, 05:21
I dont see why shouldn't it be a simple 8c2.

i think u are all wrong
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Re: PS: teams [#permalink]

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26 Jun 2008, 06:25
Jallenmorris, u rock..!...

nice explanation man...!!...
Re: PS: teams   [#permalink] 26 Jun 2008, 06:25
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# A group of 8 friends want to play doubles tennis. How many

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