NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 13:56 It is currently 26 Apr 2024, 13:56
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Four couples are to select a group that consists of three

User avatar
arjtryarjtry
Current Student
Joined: 11 May 2008
Posts: 377
Own Kudos [?]: 1029 [0]
Given Kudos: 0
Concentration: General
 Q50  V34
Send PM
Four couples are to select a group that consists of three [#permalink] New post   02 Sep 2008, 20:11
. Four couples are to select a group that consists of three people from different couples. How many such groups are possible?
(A) 24
(B) 28
(C) 32
(D) 36
(E) 40



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
asdert
Manager
Manager
Joined: 09 Oct 2007
Posts: 241
Own Kudos [?]: 372 [0]
Given Kudos: 1
Send PM
Re: Four couples are to select a group that consists of three [#permalink] New post   02 Sep 2008, 20:18
My take: (8 x 6 x 4)/3! = 32 = C

What's the OA?
User avatar
x2suresh
Director
Director
Joined: 07 Nov 2007
Posts: 718
Own Kudos [?]: 3077 [0]
Given Kudos: 5
Location: New York
Send PM
Re: Four couples are to select a group that consists of three [#permalink] New post   02 Sep 2008, 20:19
arjtryarjtry wrote:
. Four couples are to select a group that consists of three people from different couples. How many such groups are possible?
(A) 24
(B) 28
(C) 32
(D) 36
(E) 40


= 8C1 * 6C1*4C1 / 3!
= 32
User avatar
arjtryarjtry
Current Student
Joined: 11 May 2008
Posts: 377
Own Kudos [?]: 1029 [0]
Given Kudos: 0
Concentration: General
 Q50  V34
Send PM
Re: Four couples are to select a group that consists of three [#permalink] New post   02 Sep 2008, 20:24
ans is 32. but both pls provide expln as to how you came at the ans.
for eg . why is it 8c1 and not 8c3 etc etc.
the detailed working is preferred
User avatar
icandy
Director
Director
Joined: 05 Jul 2008
Posts: 623
Own Kudos [?]: 1953 [0]
Given Kudos: 1
 Q49  V41
Send PM
Re: Four couples are to select a group that consists of three [#permalink] New post   02 Sep 2008, 20:28
Total number of ways 3 people can be picked from 8 is 8 c 3 = 56

need to find possibilities in such a way that no couple is on the team

Lets find the possibilities that a couple is on the team. Any of the 4 couples can be on the team. One other person can be any one from the remaining 6 So total number of such possibilities = 4X6 =24

Reqd answer =56-24 =32
User avatar
x2suresh
Director
Director
Joined: 07 Nov 2007
Posts: 718
Own Kudos [?]: 3077 [0]
Given Kudos: 5
Location: New York
Send PM
Re: Four couples are to select a group that consists of three [#permalink] New post   02 Sep 2008, 21:15
arjtryarjtry wrote:
ans is 32. but both pls provide expln as to how you came at the ans.
for eg . why is it 8c1 and not 8c3 etc etc.
the detailed working is preferred

F1M1
F2M2
F3M3
F4M4

= Select any person from 8 members(8C1 ways) * select another person (exclude the spouse of previous person) 6C1 * select third person from the remaining(exclude spouses of first and second persons) 4C1 / 3!

F1F2F3 -- can be arranged in 3! ways
F1F2F3 - F2F1F3 - F1F3F2 - F2F3F1 -F3F1F2 - F3F2F1 .. (all these are same)
So to avoid duplicates divide by 3!.

is it clear now.
User avatar
Twoone
Intern
Intern
Joined: 02 Sep 2008
Posts: 17
Own Kudos [?]: [0]
Given Kudos: 0
Send PM
Re: Four couples are to select a group that consists of three [#permalink] New post   03 Sep 2008, 11:55
Yep,

I am also getting 32 as answer.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: Four couples are to select a group that consists of three [#permalink]
03 Sep 2008, 11:55
Moderator:
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions