Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 Jun 2016, 06:58

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
Manager
Joined: 23 Mar 2008
Posts: 219
Followers: 3

Kudos [?]: 113 [0], given: 0

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

Show Tags

08 Jun 2008, 12:03
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Joined: 17 Nov 2007
Posts: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 485

Kudos [?]: 2802 [0], given: 359

Show Tags

08 Jun 2008, 19:36
Expert's post
$$N=\frac{C^8_2*C^6_2*C^4_2*C^2_2}{P^4_4}=\frac{4*7*3*5*2*3*1}{4*3*2*1}=105$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Intern
Joined: 04 May 2007
Posts: 11
Followers: 0

Kudos [?]: 3 [0], given: 0

Show Tags

08 Jun 2008, 22:21
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
Joined: 17 Nov 2007
Posts: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 485

Kudos [?]: 2802 [0], given: 359

Show Tags

08 Jun 2008, 22:26
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)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

VP
Joined: 18 May 2008
Posts: 1287
Followers: 14

Kudos [?]: 304 [0], given: 0

Show Tags

08 Jun 2008, 22:30
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: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 485

Kudos [?]: 2802 [0], given: 359

Show Tags

08 Jun 2008, 22:33
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.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Senior Manager
Joined: 29 Aug 2005
Posts: 283
Followers: 2

Kudos [?]: 39 [0], given: 0

Show Tags

08 Jun 2008, 22:54
Pretty good explanation Walker
_________________

The world is continuous, but the mind is discrete

Current Student
Joined: 28 Dec 2004
Posts: 3385
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 14

Kudos [?]: 239 [0], given: 2

Show Tags

09 Jun 2008, 12:55
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..
Manager
Joined: 14 Mar 2008
Posts: 127
Schools: Chicago R2
Followers: 1

Kudos [?]: 19 [0], given: 0

Show Tags

13 Jun 2008, 08:09
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, 08:09
Display posts from previous: Sort by