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

It is currently 17 Sep 2014, 10:17

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

Classic Comb/Perm/Ps

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1128
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 24 [0], given: 0

GMAT Tests User
Classic Comb/Perm/Ps [#permalink] New post 08 Sep 2004, 17:27
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
A committee of 6 is chosen from 8 men and 5
women so as to contain at least 2 men and 3
women. How many different committees could
be formed if two of the men refuse to serve
together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4318
Followers: 22

Kudos [?]: 163 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:07
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700
_________________

Best Regards,

Paul

Director
Director
avatar
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 22 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:23
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700


Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.
Director
Director
User avatar
Joined: 16 Jun 2004
Posts: 893
Followers: 1

Kudos [?]: 9 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:37
Yep 635. It would be

6C1*2C1*5C4 + 6C2*5C4 + 6C2*2C1*5C3 + 6C3*5C3 =>60+75+300+200 = 635
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5097
Location: Singapore
Followers: 19

Kudos [?]: 143 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:38
Combinations can be 2m,4w
3m,3w
Since the least number of men is 2, and the least number of women is 3

For 2m,4w combinations..

Number of possible outcomes for having 4 women chosen from a group of 5 women = 5C4 = 5
Number of possible outcomes for having 2 men chosen from a group of 8 men = 8C2 = 28
But 1 pair refuse to serve together, we we're left with 28-1 = 27 pairs of men

So total number of possible outcomes for 2m4w combinations = 27*5 = 135

For 3m,3w combinations...
Number of possible outcomes for having 3 women chosen from a group of 5 women = 5C3 = 10
Number of possible outcomes for having 3 men chosen from a group of 8 men = 8C3 = 56
But 1 pair of men are fickled-minded, and so we have to remove 6C1 = 6 combinations, to give 56-6 = 50 pairs of men

Total number of possible outcomes for 3m3w combinations = 50*10 = 500

Total = 500+135 = 635
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5097
Location: Singapore
Followers: 19

Kudos [?]: 143 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:40
And as a matter of interest, if you're interested in probability, then
Number of possible outcomes = 13C6 = 1716
Number of favorable outcomes = 635

So P(at least 2 men and 3 women, 2 men refusing to serve on the same team) = 635/1716
GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4318
Followers: 22

Kudos [?]: 163 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:51
hardworker_indian wrote:
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700


Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence... :oops:
_________________

Best Regards,

Paul

Director
Director
avatar
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 22 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 18:55
Paul wrote:
hardworker_indian wrote:
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700


Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence... :oops:


Oh oh, stop staring man. You seem to do this everyday evening in the beach. :lol:
I bet someone passed by when you skipped the last senetence.
GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4318
Followers: 22

Kudos [?]: 163 [0], given: 0

GMAT Tests User
 [#permalink] New post 08 Sep 2004, 19:01
:lol: I think this is going to mark me in my GMAT quest. intr3pid what have you done to me :roll:
_________________

Best Regards,

Paul

  [#permalink] 08 Sep 2004, 19:01
    Similar topics Author Replies Last post
Similar
Topics:
4 Experts publish their posts in the topic Modifier Classic WaterFlowsUp 9 06 Aug 2014, 21:26
1 The below is a classic example of distinguishing the vivgmat 7 29 Apr 2011, 01:56
classical music puma 3 11 Apr 2008, 07:36
Experts publish their posts in the topic classic academic underachiever syndrome jaytown 3 03 May 2005, 18:06
The diet of the ordinary Greek in classical times was dipaksingh 2 16 Feb 2005, 07:23
Display posts from previous: Sort by

Classic Comb/Perm/Ps

  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®.