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

It is currently 15 Apr 2014, 20:16

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

gc test 2

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
Joined: 12 Mar 2009
Posts: 318
Followers: 1

Kudos [?]: 27 [0], given: 1

GMAT ToolKit User GMAT Tests User
gc test 2 [#permalink] New post 31 Jul 2009, 22:47
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

33% (00:00) correct 66% (00:46) wrong based on 3 sessions
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

Oa is
[Reveal] Spoiler:
c


pl explain
Senior Manager
Senior Manager
Joined: 25 Jun 2009
Posts: 315
Followers: 2

Kudos [?]: 63 [0], given: 6

GMAT Tests User
Re: gc test 2 [#permalink] New post 01 Aug 2009, 01:46
vaivish1723 wrote:
If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

Oa is
[Reveal] Spoiler:
c


pl explain


Total no. of combinations = 8C3= 56
Now lets take the case when we have one sibling in the committee. Say 1 pair then the no. of combination 6C1 X 4= 24 ( as we have 4 pair of siblings)

No. of ways when we don't have siblings in it = 56-24= 32
Manager
Manager
Joined: 14 Apr 2008
Posts: 52
Followers: 0

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

Re: gc test 2 [#permalink] New post 01 Aug 2009, 04:45
4c3 for selececting 3 couples then for every three v have 2 chioces
4c3*2*2*2
Intern
Intern
Joined: 17 Jan 2010
Posts: 6
Followers: 1

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

Re: gc test 2 [#permalink] New post 26 Jan 2010, 04:47
Why are we multiplying 4 pair of siblings by 6C1?
Senior Manager
Senior Manager
Joined: 25 Jun 2009
Posts: 315
Followers: 2

Kudos [?]: 63 [0], given: 6

GMAT Tests User
Re: gc test 2 [#permalink] New post 26 Jan 2010, 09:56
bhumika wrote:
Why are we multiplying 4 pair of siblings by 6C1?



We are trying to calculate the no. of ways in which there is one pair and one other.

so there are 6C1 ways of choosing one guy who can be paired with one pair.

Now there are 4 pairs and total no. of ways = 6C1 x 4

I hope that helps..!
Manager
Manager
Joined: 02 Oct 2009
Posts: 197
Followers: 1

Kudos [?]: 13 [0], given: 4

GMAT Tests User
Re: gc test 2 [#permalink] New post 29 Jan 2010, 21:51
keeping it simple (b1,s1),(b2,s2),(b3,s3),(b4,s4),
1 Pair produces 2 possible wasy between (b,s) with total picks is 3; 2*2*2
Then amount 4 unique pairs it can for another 4 times for unique pairing 8*4=32... brings you back to combinatorics... being a quicker calculation then the above logic...
Intern
Intern
Joined: 23 Feb 2011
Posts: 3
Followers: 0

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

GMAT Tests User
Re: gc test 2 [#permalink] New post 08 May 2011, 06:15
First find the total number of combinations without any constraints, which is 8c3 = 56 (since we're looking to make a committee of 3 people out of 8 and order doesn't matter).

Then, find out all the ways in which you would have a sibling on the committee. Let's look at one sibling pear (Brother, Sister). The number of ways they can both get on the panel is 2c2 * 6 (the six represents the 3rd person on the committee, as there are 6 people left to choose from for the last spot), which gets you 6 combinations. Multiply that 6 by 4 to incorporate the 4 different pairs of siblings.

Using the info we've calculated, the total number of combinations is 56 - 24 = 32.
Re: gc test 2   [#permalink] 08 May 2011, 06:15
    Similar topics Author Replies Last post
Similar
Topics:
Sticky, new posts 144 Experts publish their posts in the topic GMAT Club Tests 14 Math + 6 Verbal - Free for GC Members bb 152 16 Jul 2008, 09:30
New posts Percentile ranking on GC tests? TheSituation 0 20 Feb 2010, 12:15
New posts 1 To Those Grinding away on GC Tests. TheSituation 2 24 Mar 2010, 19:50
New posts Table in GC Test manulath 0 07 Jun 2012, 01:08
New posts Experts publish their posts in the topic From 630 to 750, new diagnostic test, and 20 free GC tests bb 0 29 Aug 2013, 10:05
Display posts from previous: Sort by

gc test 2

  Question banks Downloads My Bookmarks Reviews Important topics  


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®.