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

It is currently 23 Nov 2014, 21:37

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Combinatorics Question from Downloaded Materials

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 11 Mar 2007
Posts: 69
Followers: 1

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

Combinatorics Question from Downloaded Materials [#permalink] New post 03 Jun 2007, 15:48
Could someone help me on this? It is a problem included in some of the review materials I've downloaded. The answer is 80, I'm sure of that, but I'm not confident of how I got the answer.

There are 5 married couples and a group of 3 is to be formed out of them. How many arrangements are there if a married couple may not be in the same group?

I'm thinking it's easiest to calculate all of the arrangements and then subtract out the threesomes with a married couple (get your mind out of the gutter). So here goes...

(10!)/(3!)*(7!)=120 different arrangements of the 5 married couples...

If we have a married couple seated, there are (8!)/(1!)*(7!)=8 (or since there is only 1 spot left to fill and 8 people who can fill it) different combinations for one couple.

Since there are 5 couples, 5x8=40...

So 120-40=80 possible combinations.

Thanks for your help in advance.
Senior Manager
Senior Manager
User avatar
Joined: 04 Mar 2007
Posts: 447
Followers: 1

Kudos [?]: 14 [0], given: 0

 [#permalink] New post 04 Jun 2007, 03:36
I have 2 other methods of calculating:

1) 1C5*2C4 + 1C5*2C4 + 3C5 + 3C5 = 80
1C5*2C4 - one man out of 5 in combination with 2 women out of 4

1C5*2C4 - one woman out of 5 in combination with 2 men out of 4

3C5 + 3C5 - 3 men out of 5 and 3 women out of 5

2) Quantity of combinations of 3 couples out of 5 couples, multiplied by 8
8*3C5 = 80
The last method is the easiest one, but I don't understand it.
Could anyone explain?
  [#permalink] 04 Jun 2007, 03:36
    Similar topics Author Replies Last post
Similar
Topics:
3 Experts publish their posts in the topic General - Combinatorics Question microair 5 09 Nov 2010, 08:55
GMAT Question-Combinatorics 2 gmatjon 1 22 Sep 2009, 17:30
1 GMAT Question-Combinatorics gmatjon 2 22 Sep 2009, 17:25
Tough Combinatoric Questions GMAT100 12 14 Feb 2007, 20:08
Materials for download Praetorian 0 01 Jul 2005, 02:36
Display posts from previous: Sort by

Combinatorics Question from Downloaded Materials

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

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®.