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

It is currently 19 Aug 2014, 23:26

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 6 is chosen from 8 men and 5 women so as to

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
GMAT 2: 660 Q V
GPA: 3.64
WE: Accounting (Accounting)
Followers: 0

Kudos [?]: 55 [0], given: 75952

A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 15 Nov 2010, 06:47
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

59% (02:28) correct 41% (01:39) wrong based on 122 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?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510
[Reveal] Spoiler: OA
Expert Post
5 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4664
Location: Pune, India
Followers: 1069

Kudos [?]: 4771 [5] , given: 163

Re: Commitee [#permalink] New post 15 Nov 2010, 07:58
5
This post received
KUDOS
Expert's post
Geronimo wrote:
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?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510


Method 2:

8 Men - 5 Women
Choose 6 ( two cases)

Case 1:2 men, 4 women
No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2 - 1 ways.
No. of ways to choose 4 women out of 5 is 5C4.
Number of ways of choosing 2 men and 4 women = (8C2 - 1)*(5C4) = 27*5 = 135

Case 2: 3 men, 3 women
No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3 - 6C1 ways.
No of ways to choose 3 women out of 5 is 5C3.
Number of ways of choosing 3 men and 3 women = (8C3 - 6C1)*(5C3) = 500

Total number of ways = 135 + 500 = 635

Case 2: 3 men, 3 women)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19020
Followers: 3358

Kudos [?]: 24331 [2] , given: 2676

Re: Commitee [#permalink] New post 15 Nov 2010, 06:55
2
This post received
KUDOS
Expert's post
Geronimo wrote:
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?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: A.
_________________

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

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4664
Location: Pune, India
Followers: 1069

Kudos [?]: 4771 [2] , given: 163

Re: Commitee [#permalink] New post 23 Jul 2012, 21:22
2
This post received
KUDOS
Expert's post
alphabeta1234 wrote:


Hey Bunuel,

Sorry I am in the same camp as Apex. I am trying two reconcile considering that I can pick 2 men, 3 women, and 1 of the remaining, instead of having to break it into two scenarios of (2 Men, 4 Women) and (3 Men, 4 Women)

Ways to chose 6 members committee without restriction: C^2_8*C^3_5*C^1_8 =2240


What I am doing wrong in the method above? The two should give the same number.


Be careful when you do not have to arrange the people but need to choose multiple times from the same group.

Say, I want to make a team of 2 people from a group of 4 (A, B, C, D).
In how many ways can I do it?
Method1: 4C2 = 6 (AB, BC, CD, AC, AD, BD)
Method 2: If instead, I select 1 and then another, this is how I will do it: 4C1 * 3C1 = 12 (AB, BA, AC, CA...)
In the second case, I have arranged the 2 people in 2 ways. I first select A and then B in one case. I first select B and then A in another case.
But a team doesn't need arrangement, it only needs selection.
Hence, the second method is incorrect.

Similarly, in this question, you need to make a committee i.e. only select, not arrange. When you pick 2 men, 3 women and 1 of the remaining, you are double counting.

When you select 2 men in 8C2 ways, say you select A and B
When you select 3 women in 5C3 ways, say you select X, Y and Z.
When you select the last person, say you select a man C.

Now consider another case:
When you select 2 men in 8C2 ways, say you select A and C
When you select 3 women in 5C3 ways, say you select X, Y and Z.
When you select the last person, say you select a man B.

The committee is same in both the cases (A, B, C, X, Y, Z) but we count them as 2 different cases when we use your method.
Hence, there is double counting in your method. Therefore, you need to do what Bunuel suggested.

For more on P&C, check my blog: http://www.veritasprep.com/blog/categor ... er-wisdom/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19020
Followers: 3358

Kudos [?]: 24331 [1] , given: 2676

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 19 Feb 2012, 23:54
1
This post received
KUDOS
Expert's post
Apex231 wrote:
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?

Number of ways to choose 2 men = 8C2 = 28
Number of ways to choose 3 women = 5C3 = 10
Total ways to choose 2 men and 3 women = 28 * 10 = 280

Now the remaining committee member can be chosen from the 8 remaining men and women.
Total ways to choose 6 committee members = 280 * 8 = 2240

When the two men who refuse to serve together are selected -
Number of ways to choose 3 women = 5C3 = 10
Remaining 1 committee member can be chosen from the 8 remaining men and women.
Total combinations = 10 * 8 = 80


Total different committees if two of the men refuse to serve together = 2240 - 80 = 2160


I am doing something wrong here, where am i going wrong?


-----------------------

Realize my mistake, i am duplicating count when i multiply by 8.


The numbers you get will have duplications.

Let's take for example the red part of your solution above: consider five women: {A, B, C, D, E}. When you choose 3 of them (with C^3_5) you can get for example the group {A, B, C} next when you choose one from 8 people then you can get one more woman, for example D, so you'll have in the group 4 women {A, B, C, D}. Now, if you choose the group {A, B, D}, with C^3_5 and then choose C from 8 people then you'll basically will get the same 4 women group: {A, B, C, D}.

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

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 19 Apr 2010
Posts: 216
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 17 [0], given: 28

GMAT Tests User
Re: Commitee [#permalink] New post 17 Nov 2010, 06:16
VeritasPrepKarishma wrote:

Method 2:

8 Men - 5 Women
Choose 6 ( two cases)

Case 1:2 men, 4 women
No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2 - 1 ways.
No. of ways to choose 4 women out of 5 is 5C4.
Number of ways of choosing 2 men and 4 women = (8C2 - 1)*(5C4) = 27*5 = 135

Case 2: 3 men, 3 women
No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3 - 6C1 ways.
No of ways to choose 3 women out of 5 is 5C3.
Number of ways of choosing 3 men and 3 women = (8C3 - 6C1)*(5C3) = 500

Total number of ways = 135 + 500 = 635

Case 2: 3 men, 3 women)


Karishma I am not able to get the highlighted portion. Can you pls simplyfy for me.
Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4664
Location: Pune, India
Followers: 1069

Kudos [?]: 4771 [0], given: 163

Re: Commitee [#permalink] New post 17 Nov 2010, 07:23
Expert's post
prashantbacchewar wrote:
VeritasPrepKarishma wrote:

Method 2:

8 Men - 5 Women
Choose 6 ( two cases)

Case 1:2 men, 4 women
No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2 - 1 ways.
No. of ways to choose 4 women out of 5 is 5C4.
Number of ways of choosing 2 men and 4 women = (8C2 - 1)*(5C4) = 27*5 = 135

Case 2: 3 men, 3 women
No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3 - 6C1 ways.
No of ways to choose 3 women out of 5 is 5C3.
Number of ways of choosing 3 men and 3 women = (8C3 - 6C1)*(5C3) = 500

Total number of ways = 135 + 500 = 635

Case 2: 3 men, 3 women)


Karishma I am not able to get the highlighted portion. Can you pls simplyfy for me.


Sure Prashant. We are looking at the case where we take 3 men and 3 women.
Now how do we choose 3 men? If there were no constraints, it would simply be 8C3.
But, there are two men who do not want to be chosen together. Let us find out the opposite, i.e. in how many ways can we choose those two men together. Once we get this number, we can subtract it from 8C3 to get the number of ways of choosing 3 men out of 8 with the required constraints.
Now, in how many ways can we choose those two men together.
We take those two men and choose one more out of the remaining 6 using 6C1. Now we have chosen 3 men.
So 8C3 - 6C1 gives the number of ways of choosing three men when those two men are not taken together.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
GMAT 2: 660 Q V
GPA: 3.64
WE: Accounting (Accounting)
Followers: 0

Kudos [?]: 55 [0], given: 75952

Re: Commitee [#permalink] New post 18 Nov 2010, 06:52
This is what I still need to improve...
Thx for both answers !! :)
Current Student
avatar
Joined: 10 Jan 2010
Posts: 193
Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
GMAT 1: Q V
GPA: 3
WE: Consulting (Telecommunications)
Followers: 2

Kudos [?]: 22 [0], given: 7

GMAT Tests User
Re: A committee of 6 [#permalink] New post 19 Feb 2012, 06:30
Well, i do not get the 635 but narrowed it down to answer E

Total (favourable) commitees = All possible commitees - restriction

Option 1: M M W W W W (2 Men & 4 Women)
Option 2: M M M W W W (3 Men & 3 Women)
Restriction: 2 out of 8 Men refuse to serve together

Outcomes Option 1: \frac{8!}{2!*6!} * \frac{5!}{4!*1!} = 28 * 5 = 140

Outcomes Option 2: \frac{8!}{3!*5!} * \frac{5!}{3!*2!} = 56 * 10 = 560

Total commitees = 700

Outcomes restriction option 1: 1 (if 2 men refuse to serve together and 2 places are available) * 5 (Outcomes women) = 5
Outcomes restriction option 2: 3 (if 2 men refuse to serve together and 3 places are available) * 10 (Outcomes women = 30

So it should be definitely less than 700 but i guess i made a mistake somewhere. POE E
Current Student
avatar
Joined: 10 Jan 2010
Posts: 193
Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
GMAT 1: Q V
GPA: 3
WE: Consulting (Telecommunications)
Followers: 2

Kudos [?]: 22 [0], given: 7

GMAT Tests User
Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 19 Feb 2012, 07:50
Thanks Bunuel, i have found my mistake. I did not consider that the space left (at the restriction part) can be filled with all 6 remaining "men". I did just consider the possibilities to fill the places. Silly!!
Manager
Manager
avatar
Joined: 03 Oct 2009
Posts: 64
Followers: 0

Kudos [?]: 26 [0], given: 8

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 19 Feb 2012, 08:00
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?

Number of ways to choose 2 men = 8C2 = 28
Number of ways to choose 3 women = 5C3 = 10
Total ways to choose 2 men and 3 women = 28 * 10 = 280

Now the remaining committee member can be chosen from the 8 remaining men and women.
Total ways to choose 6 committee members = 280 * 8 = 2240

When the two men who refuse to serve together are selected -
Number of ways to choose 3 women = 5C3 = 10
Remaining 1 committee member can be chosen from the 8 remaining men and women.
Total combinations = 10 * 8 = 80

Total different committees if two of the men refuse to serve together = 2240 - 80 = 2160


I am doing something wrong here, where am i going wrong?


-----------------------

Realize my mistake, i am duplicating count when i multiply by 8.
Manager
Manager
avatar
Joined: 03 Oct 2009
Posts: 64
Followers: 0

Kudos [?]: 26 [0], given: 8

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 20 Feb 2012, 07:36
Thanks Bunuel for explaining..it's clear to me now..
Manager
Manager
avatar
Joined: 12 Feb 2012
Posts: 107
Followers: 1

Kudos [?]: 10 [0], given: 28

Re: Commitee [#permalink] New post 22 Jul 2012, 19:19
Bunuel wrote:
Geronimo wrote:
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?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together) : C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: A.



Hey Bunuel,

Sorry I am in the same camp as Apex. I am trying two reconcile considering that I can pick 2 men, 3 women, and 1 of the remaining, instead of having to break it into two scenarios of (2 Men, 4 Women) and (3 Men, 4 Women)

Ways to chose 6 members committee without restriction: C^2_8*C^3_5*C^1_8 =2240


What I am doing wrong in the method above? The two should give the same number.
Manager
Manager
avatar
Joined: 26 Dec 2011
Posts: 117
Followers: 1

Kudos [?]: 9 [0], given: 17

Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 25 Jul 2012, 03:23
Elegant solutions provided above; however, please let me know where did I go wrong.

8 Men, can be considered as 6men and 2men, further we have 5 women to choose from:

Thus, at least 2 and 3 mean;

Case 1: 2 men and 4 women: 6C1*2C1*5C4 + 6C2*5C4

Case 2: 3 Men and 3 Women: 6C2*2C1*5C3 + 6C3*5C3

I don't the same number of ways. What am I doing wrong?
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 67

Kudos [?]: 497 [0], given: 43

GMAT Tests User
Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 25 Jul 2012, 05:15
pavanpuneet wrote:
Elegant solutions provided above; however, please let me know where did I go wrong.

8 Men, can be considered as 6men and 2men, further we have 5 women to choose from:

Thus, at least 2 and 3 mean;

Case 1: 2 men and 4 women: 6C1*2C1*5C4 + 6C2*5C4

Case 2: 3 Men and 3 Women: 6C2*2C1*5C3 + 6C3*5C3

I don't the same number of ways. What am I doing wrong?


You are absolutely right. Why do you think you are wrong?
Check the expressions, they will give you the correct answer.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Intern
Intern
avatar
Joined: 08 Mar 2013
Posts: 19
Followers: 0

Kudos [?]: 1 [0], given: 7

Re: Commitee [#permalink] New post 23 May 2013, 19:28
Bunuel wrote:
Geronimo wrote:
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?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: A.


Been trying to figure this one out, but I'm not getting your calculations when choosing it with the two men together. How do you arrive at C^2_2 and the (C^2_2*C^1_6) . Sorry if I'm being a bit slow on this one, but just haven't wrapped my brain around this.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19020
Followers: 3358

Kudos [?]: 24331 [0], given: 2676

Re: Commitee [#permalink] New post 24 May 2013, 00:39
Expert's post
Dixon wrote:
Bunuel wrote:
Geronimo wrote:
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?

(1) 635
(2) 700
(3) 1404
(4) 2620
(5) 3510


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: A.


Been trying to figure this one out, but I'm not getting your calculations when choosing it with the two men together. How do you arrive at C^2_2 and the (C^2_2*C^1_6) . Sorry if I'm being a bit slow on this one, but just haven't wrapped my brain around this.


2 men and 4 women: C^2_2*C^4_5=1*5. One way to choose 2 particular men out of 2: C^2_2=1.

3 men and 3 women: (C^2_2*C^1_6)*C^3_5. One way to choose 2 particular men out of 2: C^2_2=1 and 6 ways to choose the third men out of the remaining 6: C^1_6.

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

Get the best GMAT Prep Resources with GMAT Club Premium Membership

SVP
SVP
User avatar
Joined: 09 Sep 2013
Posts: 2064
Followers: 178

Kudos [?]: 34 [0], given: 0

Premium Member
Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 12 Jul 2014, 14:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: A committee of 6 is chosen from 8 men and 5 women so as to   [#permalink] 12 Jul 2014, 14:55
    Similar topics Author Replies Last post
Similar
Topics:
A Committee of 6 is chosen from 8 men and 5 women, so as to neelesh 1 25 Dec 2007, 11:56
A Committee of 6 is chosen from 8 men and 5 women, so as to chillpill 4 16 Apr 2006, 18:08
A Committee of 6 is chosen from 8 men and 5 women, so as to mbassmbass04 7 08 Jan 2005, 17:34
A committee of 6 is chosen from 8 men and 5 women so as to rahul 6 10 Sep 2004, 01:43
A committee of 6 is chosen from 8 men and 5 women so as to Bhai 5 29 Aug 2004, 18:22
Display posts from previous: Sort by

A committee of 6 is chosen from 8 men and 5 women so as to

  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®.