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

It is currently 20 Oct 2019, 02:23

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 people is to be chosen from the president

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

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 12 Oct 2011
Posts: 106
GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 08 Apr 2012, 06:43
8
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

62% (01:59) correct 38% (02:04) wrong based on 138 sessions

HideShow timer Statistics

A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58443
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 08 Apr 2012, 09:21
BN1989 wrote:
A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40


Each company can send only one "representative" to the committee. Let's see in how many ways we can choose 3 companies (as there should be 3 members) to send only one "representatives" to the committee: 4C3=4.

But these 3 chosen companies can send two persons (either president or vice president ): 2*2*2=2^3=8.

Total # of ways: 4C3*2^3=32.

Answer: D.

Or: 8*6*4/3!=32, we are dividing by 3! because the order in the committee is not important.

Answer: D.

Similar problems with different approaches:
combination-permutation-problem-couples-98533.html
ps-combinations-94068.html
committee-of-88772.html

Hope it helps.
_________________
Manager
Manager
avatar
Joined: 03 Aug 2011
Posts: 209
Location: United States
Concentration: General Management, Entrepreneurship
GMAT 1: 750 Q49 V44
GPA: 3.38
WE: Engineering (Computer Software)
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post Updated on: 08 Apr 2012, 12:26
man bunuel always has such a quick and elegant way. Can you comment if this approach is correct?

i just did the total number of possibilities first:

ab cd ef gh are the 4 companies for example

8c3. this will include the problem counts of a pres/vp of the same company

how many ways did we over count = How many ways can a president and VP of the same company be part of this group?

choose the group, 4c1, then take both the president and vp, 2c2, then a remaining member from one of the other 3, 6c1

4c1 * 2c2 * 6c1
4 * 1 * 6 = 24

8c3 - 24 = 56 - 24 = 32

Originally posted by pinchharmonic on 08 Apr 2012, 11:16.
Last edited by pinchharmonic on 08 Apr 2012, 12:26, edited 1 time in total.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58443
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 08 Apr 2012, 11:20
pinchharmonic wrote:
man bunuel always has such a quck and elegant way. Can you comment if this approach is correct?

i just did the total number of possibilities first:

ab cd ef gh are the 4 companies for example

8c3. this will include the problem counts of a pres/vp of the same company

how many ways did we over count = How many ways can a president and VP of the same company be part of this group?

choose the group, 4c1, then take both the president and vp, 2c2, then a remaining member from one of the other 3, 6c1

4c1 * 2c2 * 6c1
4 * 1 * 6 = 24

8c3 - 24 = 56 - 24 = 32


Yes, that's also a correct approach.
_________________
Intern
Intern
User avatar
Joined: 06 Aug 2011
Posts: 1
Location: India
Concentration: General Management, Sustainability
WE: Engineering (Telecommunications)
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 02 Jul 2015, 05:05
Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58443
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 02 Jul 2015, 05:40
pptkrishanth wrote:
Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.


Yes. This is the same exact approach given in my first reply above.
_________________
Director
Director
avatar
G
Joined: 09 Mar 2018
Posts: 994
Location: India
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 16 Feb 2019, 08:59
BN1989 wrote:
A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40


Now this is quite amusing, i would consider the P & VP to be a couple :-D

So there will be 4 couples, out which we need a committee which wont have one

Total number of ways, without any restriction =\(5C_3\) = 56 ways

Now if we keep a restriction we can calculate the ways in which one couple is present = \(4C_1 * 6C_1\)= 24 ways

Now the difference will give us 32 ways
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Intern
Intern
avatar
B
Joined: 13 Jun 2018
Posts: 37
GMAT 1: 700 Q49 V36
Re: A committee of 3 people is to be chosen from the president  [#permalink]

Show Tags

New post 16 Feb 2019, 09:52
mygmatsuccess wrote:
Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.



I understand that we are dividing by 3! because the order does not matter. Can someone please explain the math behind it?

Bunuel
GMAT Club Bot
Re: A committee of 3 people is to be chosen from the president   [#permalink] 16 Feb 2019, 09:52
Display posts from previous: Sort by

A committee of 3 people is to be chosen from the president

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





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