GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Oct 2018, 17:57

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.

Close

Request Expert Reply

Confirm Cancel

A committee of 3 men and 3 women must be formed from a group of 6 men

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4200
A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 02 Feb 2005, 20:11
3
27
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

37% (02:18) correct 63% (02:19) wrong based on 82 sessions

HideShow timer Statistics

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?

(A) 1120
(B) 910
(C) 810
(D) 560
(E) 210

_________________

Best Regards,

Paul

Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49915
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 02 Sep 2013, 01:50
6
6
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 the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

General Discussion
Director
Director
avatar
Joined: 19 Nov 2004
Posts: 529
Location: SF Bay Area, USA
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 02 Feb 2005, 22:51
= Total combination - combination in which the man and women serve together
6c3*8c3 - 5c2*7c2
Intern
Intern
avatar
Joined: 28 Dec 2004
Posts: 32
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 05 Mar 2005, 19:52
Hi,

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

Thanks,

Mike
Director
Director
avatar
Joined: 18 Feb 2005
Posts: 642
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 05 Mar 2005, 21:11
2
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
VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1083
Location: London, UK
Schools: Tuck'08
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 06 Mar 2005, 07:37
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
avatar
Joined: 30 Sep 2004
Posts: 1439
Location: Germany
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 06 Mar 2005, 07:40
1
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).
Manager
Manager
avatar
Joined: 27 Jan 2005
Posts: 97
Location: San Jose,USA- India
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 09 Mar 2005, 00:17
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
avatar
Joined: 30 May 2013
Posts: 160
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
GMAT ToolKit User
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 01 Sep 2013, 22:15
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
V
Joined: 02 Sep 2009
Posts: 49915
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 02 Sep 2013, 01:53
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 the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 13 Dec 2013
Posts: 37
GMAT 1: 620 Q42 V33
GMAT ToolKit User
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 19 Apr 2014, 19:48
Hi,

I tried to do it the other way around, instead of: total combos - combos restricted, I tried the approach of adding up all permited combos.

Combos where the man is but the woman is left out: 6C3*7C3 (Only take 7 women into account, not 8)
Combos where the woman is but the man is left out: 5C3*8C3 (Only take 5 men into account, not 6)
Combos where neither is in a group selected: 5C3*7C3 (Both are taken out)

Adding up these three scenarios, I get a total of 1,530 combos.

Could someone help me out here? Can't seem to understand where i'm over estimating.

Much appreciated.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49915
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 20 Apr 2014, 03:08
2
Enael wrote:
Hi,

I tried to do it the other way around, instead of: total combos - combos restricted, I tried the approach of adding up all permited combos.

Combos where the man is but the woman is left out: 6C3*7C3 (Only take 7 women into account, not 8)
Combos where the woman is but the man is left out: 5C3*8C3 (Only take 5 men into account, not 6)

Combos where neither is in a group selected: 5C3*7C3 (Both are taken out)

Adding up these three scenarios, I get a total of 1,530 combos.

Could someone help me out here? Can't seem to understand where i'm over estimating.

Much appreciated.


The number of committees with John but not Mary: \((1*C^2_5)(C^3_7)=10*35=350\);

The number of committees with Mary but not John: \((C^3_5)(1*C^2_7)=10*21=210\);

The number of committees without John and without Mary: \((C^3_5)(C^3_7)=10*35=350\).

Total = 350 + 350 + 210 = 910.

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

CEO
CEO
User avatar
P
Joined: 08 Jul 2010
Posts: 2537
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
Re: A committee of 3 men and 3 women must be formed from a group of 6 men  [#permalink]

Show Tags

New post 20 Apr 2018, 02:21
1
Paul wrote:
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?



Please find the solution with two methods in attachment

Bunuel

Could you please add the options in question???

A) 1120
B) 910
C) 810
D) 560
E) 210
Attachments

File comment: www.GMATinsight.com
Screen Shot 2018-04-20 at 2.48.06 PM.png
Screen Shot 2018-04-20 at 2.48.06 PM.png [ 447.24 KiB | Viewed 4538 times ]


_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49915
Re: A committee of 3 men and 3 women must be formed from a group  [#permalink]

Show Tags

New post 20 Apr 2018, 02:23
GMATinsight wrote:
Paul wrote:
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?



Please find the solution with two methods in attachment

Bunuel

Could you please add the options in question???

A) 1120
B) 910
C) 810
D) 560
E) 210

_______________
Done. Thank you.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

GMAT Club Bot
Re: A committee of 3 men and 3 women must be formed from a group &nbs [#permalink] 20 Apr 2018, 02:23
Display posts from previous: Sort by

A committee of 3 men and 3 women must be formed from a group of 6 men

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

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

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.