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

It is currently 16 Apr 2014, 08:22

Hurry Up:

Last day of registration for Dealing with a Ding - Webinar by GMATClub and Admissionado Consulting.


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

A committee of 3 men and 3 women must be formed from a group

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
GMAT Club Legend
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4318
Followers: 16

Kudos [?]: 123 [0], given: 0

GMAT Tests User
A committee of 3 men and 3 women must be formed from a group [#permalink] New post 02 Feb 2005, 19:11
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

40% (02:33) correct 60% (02:17) wrong based on 5 sessions
A committee of 3 men and 3 women must be formed from a group of 6 men and 8 women. How many such committees can we form if 1 man and 1 woman refuse to serve together?
_________________

Best Regards,

Paul

2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 17283
Followers: 2867

Kudos [?]: 18328 [2] , given: 2345

GMAT Tests User CAT Tests
Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 02 Sep 2013, 00:50
2
This post received
KUDOS
Expert's post
rrsnathan wrote:
Hi Bunuel,

Can u explain this problem

in the above explanation its states that "Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2. "

If John and mary refused to work together then how can we select those people already and subtract 5C2*7C2 from total combination???

Pls explain this.

Thanks and Regards,
Rrsnathan


A committee of 3 men and 3 women must be formed from a group of 6 men and 8 women. How many such committees can we form if 1 man and 1 woman refuse to serve together?

The total # of committees without the restriction is C^3_6*C^3_8;
The # of committees which have both John and Mary is 1*1*C^2_5*C^2_7 (one way to select John, 1 way to select Mary, selecting the remaining 2 men from 5, selecting the remaining 2 women from 7).

{Total} - {Restriction} = C^3_6*C^3_8-C^2_5*C^2_7.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Director
Director
Joined: 19 Nov 2004
Posts: 567
Location: SF Bay Area, USA
Followers: 3

Kudos [?]: 20 [0], given: 0

GMAT Tests User
 [#permalink] New post 02 Feb 2005, 21:51
= Total combination - combination in which the man and women serve together
6c3*8c3 - 5c2*7c2
VP
VP
Joined: 18 Nov 2004
Posts: 1448
Followers: 2

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

GMAT Tests User
 [#permalink] New post 02 Feb 2005, 22:01
6C3x8C3 - 5C2x7C2 = 910
GMAT Club Legend
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4318
Followers: 16

Kudos [?]: 123 [0], given: 0

GMAT Tests User
 [#permalink] New post 03 Feb 2005, 07:20
:yes 910 is the OA. Too simple it is :lol:
_________________

Best Regards,

Paul

Intern
Intern
Joined: 28 Dec 2004
Posts: 35
Followers: 0

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

confused [#permalink] New post 05 Mar 2005, 18:52
Hi,

I was a little confused about how you determine: (5,2)*(7,2)?

Thanks,

Mike
Senior Manager
Senior Manager
Joined: 25 Oct 2004
Posts: 250
Followers: 1

Kudos [?]: 3 [0], given: 0

GMAT Tests User
 [#permalink] New post 05 Mar 2005, 19:29
910

no of ways= total no of ways- when they are always together
=6c3*8c3-5c2*7c2
Intern
Intern
Joined: 28 Dec 2004
Posts: 35
Followers: 0

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

not sure [#permalink] New post 05 Mar 2005, 19:33
sorry.. i'm not sure how everyone arrived at " 5c2*7c2 "

why does this equal the number of combinations that the 1 man and 1 women who can't be together? :oops:
Director
Director
Joined: 18 Feb 2005
Posts: 674
Followers: 1

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

GMAT Tests User
 [#permalink] New post 05 Mar 2005, 20:11
1 man and 1 woman are already selected so
You can select the remaining 2 men and 2 women from 5 men and 7 women

So 5C2*7C2
Senior Manager
Senior Manager
Joined: 19 Feb 2005
Posts: 493
Location: Milan Italy
Followers: 1

Kudos [?]: 7 [0], given: 0

GMAT Tests User
 [#permalink] New post 06 Mar 2005, 02:25
WOOOOOOOOOOOOOOOOOO
910!!!! That was written on my paper!
Am I dreaming?
:bouncer2
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1128
Location: London, UK
Schools: Tuck'08
Followers: 6

Kudos [?]: 19 [0], given: 0

GMAT Tests User
 [#permalink] New post 06 Mar 2005, 06:37
gmat2me2 wrote:
1 man and 1 woman are already selected so
You can select the remaining 2 men and 2 women from 5 men and 7 women

So 5C2*7C2


sorry guys, I don't get it.
I agree on the way to calculate it : total outcome - outcome when the man and the woman are together in the group
I found 6C3*8C3 - 6C1*8C1..which is wrong but i can not understand why my answer is wrong and why 5c2*7c2 is good ? 5c2*7c2 just deal with 2 people , it seems incomplete to me... please help
VP
VP
Joined: 30 Sep 2004
Posts: 1496
Location: Germany
Followers: 4

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

GMAT Tests User
 [#permalink] New post 06 Mar 2005, 06:40
Antmavel wrote:
gmat2me2 wrote:
1 man and 1 woman are already selected so
You can select the remaining 2 men and 2 women from 5 men and 7 women

So 5C2*7C2


sorry guys, I don't get it.
I agree on the way to calculate it : total outcome - outcome when the man and the woman are together in the group
I found 6C3*8C3 - 6C1*8C1..which is wrong but i can not understand why my answer is wrong and why 5c2*7c2 is good ? 5c2*7c2 just deal with 2 people , it seems incomplete to me... please help


5c2*7c2 => means that THE women and THE man is already in the group. so 4 places are left for 2 out of 5 (5c2) and 2 out of 7 (7c2).
Intern
Intern
Joined: 28 Dec 2004
Posts: 35
Followers: 0

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

thanks!! [#permalink] New post 06 Mar 2005, 06:43
Thanks! I get it now!! I should have got it before. :shock:
Manager
Manager
Joined: 27 Jan 2005
Posts: 100
Location: San Jose,USA- India
Followers: 1

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

 [#permalink] New post 08 Mar 2005, 23:17
Antmavel wrote:
I found 6C3*8C3 - 6C1*8C1..which is wrong but i can not understand why my answer is wrong and why 5c2*7c2 is good ? 5c2*7c2 just deal with 2 people , it seems incomplete to me... please help


Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2.
Manager
Manager
Joined: 30 May 2013
Posts: 142
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Followers: 0

Kudos [?]: 20 [0], given: 64

GMAT ToolKit User
Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 01 Sep 2013, 21:15
Hi Bunuel,

Can u explain this problem

in the above explanation its states that "Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2. "

If John and mary refused to work together then how can we select those people already and subtract 5C2*7C2 from total combination???

Pls explain this.

Thanks and Regards,
Rrsnathan
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 17283
Followers: 2867

Kudos [?]: 18328 [0], given: 2345

GMAT Tests User CAT Tests
Re: A committee of 3 men and 3 women must be formed from a group [#permalink] New post 02 Sep 2013, 00:53
Expert's post
Bunuel wrote:
rrsnathan wrote:
Hi Bunuel,

Can u explain this problem

in the above explanation its states that "Question says 1 man and 1 woman refuse to server. Let's say persons refuse to server are John and Mary. 6C1*8C1 would mean you are selecting any 1 man from 6 men(John may or may not be there in the selection) and any 1 woman from 8 women(Mary may or may not be there in the selection).So it is wrong.

Instead, we know John is already selected, so we are left with picking 2 men from 5,and as Mary is already selected we are left with picking 2 women from 7.Thus comes 5C2*7C2. "

If John and mary refused to work together then how can we select those people already and subtract 5C2*7C2 from total combination???

Pls explain this.

Thanks and Regards,
Rrsnathan


A committee of 3 men and 3 women must be formed from a group of 6 men and 8 women. How many such committees can we form if 1 man and 1 woman refuse to serve together?

The total # of committees without the restriction is C^3_6*C^3_8;
The # of committees which have both John and Mary is 1*1*C^2_5*C^2_7 (one way to select John, 1 way to select Mary, selecting the remaining 2 men from 5, selecting the remaining 2 women from 7).

{Total} - {Restriction} = C^3_6*C^3_8-C^2_5*C^2_7.

Hope it's clear.


Similar questions:
at-a-meeting-of-the-7-joint-chiefs-of-staff-the-chief-of-154205.html
a-committee-of-6-is-chosen-from-8-men-and-5-women-so-as-to-104859.html
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Re: A committee of 3 men and 3 women must be formed from a group   [#permalink] 02 Sep 2013, 00:53
    Similar topics Author Replies Last post
Similar
Topics:
New posts Ben needs to form a committee of 3 from a group of 8 Zem 7 06 Apr 2005, 10:04
New posts Ben needs to form a committee of 3 from a group of 8 rc1979 8 23 Apr 2005, 15:10
New posts A committee of 3 has to be formed randomly from a group of 6 MooseDrool 6 23 Aug 2007, 18:48
Popular new posts 8 Experts publish their posts in the topic A committee of 3 has to be formed randomly from a group of 6 pmal04 11 18 Jul 2009, 13:03
New posts 1 Experts publish their posts in the topic A committee of 3 has to be formed randomly from a group of 6 teal 5 20 Mar 2012, 15:19
Display posts from previous: Sort by

A committee of 3 men and 3 women must be formed from a group

  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®.