Find all School-related info fast with the new School-Specific MBA Forum
Learn more about the recent changes... close

It is currently 25 May 2013, 16:52
Customize  |  Hide

Search
Register
New posts
Unanswered

There are 5 married couples and a group of three is to be

  Question banks Downloads My Bookmarks Reviews  
  Oldest Best Reply
Search for:
 
Print view
First unread post
Author Message
TAGS:
Manager
Manager
Joined: 10 Dec 2005
Posts: 104
Followers: 1

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

There are 5 married couples and a group of three is to be [#permalink] New post 20 Dec 2005, 20:48
00:00

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.
_________________

JAI HIND!

Senior Manager
Senior Manager
Joined: 05 Oct 2005
Posts: 487
Followers: 1

Kudos [?]: 1 [0], given: 0

GMAT Tests User
Re: PS: P&C [#permalink] New post 20 Dec 2005, 21:07
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.


I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Answer: 80
VP
VP
User avatar
Joined: 14 Sep 2005
Posts: 1001
Location: South Korea
Followers: 1

Kudos [?]: 10 [0], given: 0

GMAT Tests User
 [#permalink] New post 20 Dec 2005, 21:14
Total number of cases to pick 3 = 10C3 = 120
Number of cases to pick 3 including one couple = 5C1 * 8C1 = 40

120 - 40 = 80

I guess yb's approach is way better.
_________________

Auge um Auge, Zahn um Zahn :twisted: !

SVP
SVP
Joined: 14 Dec 2004
Posts: 1714
Followers: 1

Kudos [?]: 17 [0], given: 0

GMAT Tests User
Re: PS: P&C [#permalink] New post 21 Dec 2005, 12:21
cool_jonny009 wrote:
yb wrote:
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.


I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Answer: 80


Can you explain why you divided by 3!


Please allow me....

You divide by 3! in order to get rid of same group of 3 fixed people arranged in different order & keep only one. i.e. X Y Z is actually one group but without division by 3!, all possible combinations of X Y Z will be treated as different groups.

In other words, (10*8*6) gives you total number of possible groups but this is permutation but actually we need combination, so divide by 3!
Senior Manager
Senior Manager
Joined: 05 Oct 2005
Posts: 487
Followers: 1

Kudos [?]: 1 [0], given: 0

GMAT Tests User
Re: PS: P&C [#permalink] New post 21 Dec 2005, 12:30
vivek123 wrote:
cool_jonny009 wrote:
yb wrote:
JAI HIND wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I solved this. But wanna check if you guys have a shorter method.


I'm not too good at perm/comb, but here goes:

10 ways to pick the first person, 8 ways to pick the second (b/c the 1st person's spouse must be excluded), and 6 ways to pick the 3rd (b/c the 2nd person's spouse must be excluded).

Therefore,

(10*8*6)/3! since order does not matter.

Answer: 80


Can you explain why you divided by 3!


Please allow me....

You divide by 3! in order to get rid of same group of 3 fixed people arranged in different order & keep only one. i.e. X Y Z is actually one group but without division by 3!, all possible combinations of X Y Z will be treated as different groups.

In other words, (10*8*6) gives you total number of possible groups but this is permutation but actually we need combination, so divide by 3!


Yes, I agree. To further clarify, note that the order of XYZ does not matter.
XYZ could be :

XYZ
XZY
YXZ
YZX
ZYX
ZXY
or 3! = 3*2 = 6 ways.

Therefore, we must divide by 3! since we don't care about the order.
Re: PS: P&C   [#permalink] 21 Dec 2005, 12:30
    Similar topics Author Replies Last post
Similar
Topics:
 New posts There are 5 married couples and a group of three is to be preyshi 3 06 Dec 2003, 16:32
New posts There are 5 married couples and a group of three is to be vivek_dj 6 30 Jan 2004, 18:19
New posts There are 5 married couples and a groups of three is to be dr_sabr 8 03 Jul 2004, 11:51
New posts There are 5 married couples and a group of three is to be ddpp128 4 16 Oct 2004, 13:46
Popular new posts There are 5 married couples and a group of three is to be sk 15 06 Feb 2005, 17:55
Display posts from previous: Sort by

There are 5 married couples and a group of three is to be

  Question banks Downloads My Bookmarks Reviews  
Print view
First unread post

Moderators: Bunuel, VeritasPrepKarishma, IanStewart

Search for:


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

Main navigation

GMAT Resources

Partners

MBA Resources

www.gmac.com|www.mba.com