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# A group of 8 friends wants to play double tennis. How many

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Manager
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A group of 8 friends wants to play double tennis. How many [#permalink]  08 Jun 2008, 11:03
A group of 8 friends wants to play double 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
CEO
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Kudos [?]: 2140 [0], given: 359

Re: double tennis [#permalink]  08 Jun 2008, 18:36
Expert's post
$$N=\frac{C^8_2*C^6_2*C^4_2*C^2_2}{P^4_4}=\frac{4*7*3*5*2*3*1}{4*3*2*1}=105$$
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Re: double tennis [#permalink]  08 Jun 2008, 21:21
walker wrote:
$$N=\frac{C^8_2*C^6_2*C^4_2*C^2_2}{P^4_4}=\frac{4*7*3*5*2*3*1}{4*3*2*1}=105$$

Hi Walker,

What is the logic behind 4P4 ?
CEO
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Kudos [?]: 2140 [0], given: 359

Re: double tennis [#permalink]  08 Jun 2008, 21:26
Expert's post
mandy12 wrote:
What is the logic behind 4P4 ?

When we choose four groups, it is necessary to exclude order among groups, for example:

12,34,56,78 and 34,12,56,78 is the same option. (1,2,3,4,5,6,7,8 - people)
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Re: double tennis [#permalink]  08 Jun 2008, 21:30
Can u explain the logic behind this formula?
walker wrote:
$$N=\frac{C^8_2*C^6_2*C^4_2*C^2_2}{P^4_4}=\frac{4*7*3*5*2*3*1}{4*3*2*1}=105$$
CEO
Joined: 17 Nov 2007
Posts: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Kudos [?]: 2140 [0], given: 359

Re: double tennis [#permalink]  08 Jun 2008, 21:33
Expert's post
$$C^8_2$$ - choose first team

$$C^6_2$$ - choose second team

$$C^4_2$$ - choose third team

$$C^2_2$$ - choose fourth team

$$P^4_4$$ - exclude order among four teams.
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Re: double tennis [#permalink]  08 Jun 2008, 21:54
Pretty good explanation Walker
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Re: double tennis [#permalink]  09 Jun 2008, 11:55
for some reason i get 115..

i did as follows.. [8C2*6C2*4C2*2C2]/4!

i use 4! because suppose each player is a nd B..then we have counted that pair twice i.e ab, ba etc for each of the 4 slots..
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Re: double tennis [#permalink]  13 Jun 2008, 07:09
maybe I'm not so skilled in combinative formulas but for me this is the way: for even N the quantity of pairs = (N-1)*(N-3)*...*(N-n), where n = N-1. So that for 8 players it's 7*5*3*1=105. For N=10 it's 9*7*5*3*1 = 945
P.S. I should explain how I reached that. I draw a picture with 4 lines then investigated those combinations, it was 3. Why 3, cause if we chosen 1 pair only 1 pair left. How many first pairs I could chose - 3 (N-1).
OK, how about 6? I chosen 1 pair and what I saw, right, 4 lines. I had already seen them and knew that they consisted 3 possible pairs. How many first pairs I might chose, correct, 5 (N-1). So that we had (N-5)*(N-3)*(N-1). I kept reasoning and found the regularity and wrote it in the opposite way - see above .
All this reasoning took around 1 minute...
Re: double tennis   [#permalink] 13 Jun 2008, 07:09
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