It is currently 14 Dec 2017, 00:30

Decision(s) Day!:

CHAT Rooms | Ross R1 | Kellogg R1 | Darden R1 | Tepper R1


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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

M08-04

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

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42599

Kudos [?]: 135577 [0], given: 12701

M08-04 [#permalink]

Show Tags

New post 15 Sep 2014, 23:36
Expert's post
7
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

69% (01:30) correct 31% (00:56) wrong based on 102 sessions

HideShow timer Statistics

If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

A. 36
B. 60
C. 72
D. 80
E. 100
[Reveal] Spoiler: OA

_________________

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

Kudos [?]: 135577 [0], given: 12701

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42599

Kudos [?]: 135577 [2], given: 12701

Re M08-04 [#permalink]

Show Tags

New post 15 Sep 2014, 23:36
2
This post received
KUDOS
Expert's post
Official Solution:

If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

A. 36
B. 60
C. 72
D. 80
E. 100

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is \(C_3^{10}\). From this figure we have to subtract the number of committees that consist entirely of men: \(C_3^6\). The final answer is \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E
_________________

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

Kudos [?]: 135577 [2], given: 12701

Intern
Intern
avatar
Joined: 07 Aug 2015
Posts: 5

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

Re: M08-04 [#permalink]

Show Tags

New post 17 Aug 2015, 01:02
Bunuel wrote:
Official Solution:

If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

A. 36
B. 60
C. 72
D. 80
E. 100

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is \(C_3^{10}\). From this figure we have to subtract the number of committees that consist entirely of men: \(C_3^6\). The final answer is \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E


Hello Bunuel,

I think that there is an error with the way you edited the combinaisions which are inverted meaning that you're picking 6 men out 3 (\(C_3^6\)) which is impossible (am I right?).

Also I would have liked to know if there was another way of solving this problem by "fixing" one women and then choosing two other members out of the nine remaining. I tried to do so using this formula but I didn't get the right solution:
# of ways to select one women out of four = 4
# of ways to select two other members out of nine = 36
\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

Carnivor646

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

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42599

Kudos [?]: 135577 [0], given: 12701

Re: M08-04 [#permalink]

Show Tags

New post 17 Aug 2015, 03:20
Carnivor646 wrote:
Bunuel wrote:
Official Solution:

If 4 women and 6 men work in the accounting department, in how many ways can a committee of 3 be formed if it has to include at least one woman?

A. 36
B. 60
C. 72
D. 80
E. 100

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is \(C_3^{10}\). From this figure we have to subtract the number of committees that consist entirely of men: \(C_3^6\). The final answer is \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E


Hello Bunuel,

I think that there is an error with the way you edited the combinaisions which are inverted meaning that you're picking 6 men out 3 (\(C_3^6\)) which is impossible (am I right?).



\(C^3_6\), \(C^6_3\), 6C3 are the same: choosing 3 out of 6. How can we choose 6 out of 3? Those are just different ways of writing the same.
_________________

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

Kudos [?]: 135577 [0], given: 12701

Current Student
avatar
Joined: 10 Aug 2014
Posts: 50

Kudos [?]: 9 [0], given: 2

GMAT 1: 680 Q49 V34
Reviews Badge
Re: M08-04 [#permalink]

Show Tags

New post 06 Sep 2015, 13:26
Bunuel,
i tried to solve problem by using this formula:
1C4 - choosing 1 woman
2C4 - 2 woman out 4
3C4 - 3 woman out of 4

2C6 - 2 man out of 6
1C6 - 1 man out of 6

1C4+2C6+2C4+1C6+3C4=4+15+6+6+4=35
Can you please show me where i went wrong?

Kudos [?]: 9 [0], given: 2

Intern
Intern
avatar
Joined: 07 Aug 2015
Posts: 5

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

Re: M08-04 [#permalink]

Show Tags

New post 07 Sep 2015, 01:11
Hello Bunuel,

Could you tell why my solution of "fixing" one women and then choosing two other members out of the nine remaining didn't work please? As I said, I tried to do so using this formula but my solution did not match yours:

# of ways to select one women out of four = 4
# of ways to select two other members out of nine = 36
\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

Carnivor646

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

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42599

Kudos [?]: 135577 [1], given: 12701

Re: M08-04 [#permalink]

Show Tags

New post 07 Sep 2015, 02:19
1
This post received
KUDOS
Expert's post
Carnivor646 wrote:
Hello Bunuel,

Could you tell why my solution of "fixing" one women and then choosing two other members out of the nine remaining didn't work please? As I said, I tried to do so using this formula but my solution did not match yours:

# of ways to select one women out of four = 4
# of ways to select two other members out of nine = 36
\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

Carnivor646


The number you get will have duplications.

The number you get will have duplication.

Consider the group of three women {ABC}
Say you select women A with 4C1, next you can get women B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group.
_________________

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

Kudos [?]: 135577 [1], given: 12701

Intern
Intern
avatar
Joined: 03 Jul 2015
Posts: 32

Kudos [?]: 10 [0], given: 5

Re: M08-04 [#permalink]

Show Tags

New post 08 Jan 2016, 06:55
I have another solution. Clearly something is wrong with it.
Bunuel, Could you Please tell me why ? For my understanding only .

So if we have to select 3 people from 4 women 6 men ; at least one of them has to be a woman.
We can select in 3 ways :
1 woman 2 men : 4C1 X 6C2
2 women 1 man : 4C2 X 6C1
3 women : 4C3
We ad them all : 19+36+4 = 59 :shock: :shock:

Thanks a tonne

Kudos [?]: 10 [0], given: 5

Intern
Intern
avatar
Joined: 22 Apr 2015
Posts: 12

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

GMAT 1: 760 Q50 V44
Re: M08-04 [#permalink]

Show Tags

New post 08 Jan 2016, 08:14
shreyashid wrote:
I have another solution. Clearly something is wrong with it.
Bunuel, Could you Please tell me why ? For my understanding only .

So if we have to select 3 people from 4 women 6 men ; at least one of them has to be a woman.
We can select in 3 ways :
1 woman 2 men : 4C1 X 6C2
2 women 1 man : 4C2 X 6C1
3 women : 4C3
We ad them all : 19+36+4 = 59 :shock: :shock:

Thanks a tonne


shreyashid

You're actually right, but your math is wrong.
4C1 = 4 and 6C2 = 15

So you should be adding up 60+36+4 = 100.

However, you should also note that the published solution takes the least amount of calculation (2 choose functions vs 6), meaning that there is much more room for error!

Cheers

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

Intern
Intern
avatar
Joined: 03 Jul 2015
Posts: 32

Kudos [?]: 10 [0], given: 5

Re: M08-04 [#permalink]

Show Tags

New post 08 Jan 2016, 08:21
oops!!
Thanks

Kudos [?]: 10 [0], given: 5

Intern
Intern
avatar
B
Joined: 25 Jul 2017
Posts: 2

Kudos [?]: 0 [0], given: 18

CAT Tests
Re: M08-04 [#permalink]

Show Tags

New post 14 Nov 2017, 20:37
mika84 wrote:
Bunuel,
i tried to solve problem by using this formula:
1C4 - choosing 1 woman
2C4 - 2 woman out 4
3C4 - 3 woman out of 4

2C6 - 2 man out of 6
1C6 - 1 man out of 6

1C4+2C6+2C4+1C6+3C4=4+15+6+6+4=35
Can you please show me where i went wrong?




The approach is correct but the way you applied it isn't. Note the distinction between AND and OR. Let's say we have three spaces to fill.

A B C

Case 1:

Two men "AND" one woman. Whenever we have "AND", we multiply.

So, 6C2 * 4 = 60

Case 2:

One man and two women.

6 * 4C2 = 36

Case 3:

All 3 women.

4C3 = 4

Now, we can have either of these choices. MMF "OR" MFF "OR" FFF. We add them in case of "OR". That gives us:

60+36+4 = 100

Simply put, multiply when you have AND and add when you have OR. Hope this helps.

Kudos [?]: 0 [0], given: 18

Re: M08-04   [#permalink] 14 Nov 2017, 20:37
Display posts from previous: Sort by

M08-04

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

Moderators: chetan2u, Bunuel



GMAT Club MBA Forum Home| About| Terms and Conditions| 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®.