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

It is currently 27 Jun 2016, 16:00
GMAT Club Tests

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 three-person committee must be chosen from a group of 7

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Senior Manager
Senior Manager
avatar
Joined: 03 Nov 2005
Posts: 395
Location: Chicago, IL
Followers: 3

Kudos [?]: 41 [0], given: 17

A three-person committee must be chosen from a group of 7 [#permalink]

Show Tags

New post 18 Jan 2006, 20:20
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (03:20) wrong based on 5 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

Plz, help me out with this, guys.

My solution: C(7;1)*C(16;2)=7*120=840
What's wrong with my reasoning here.
_________________

Hard work is the main determinant of success

Manager
Manager
avatar
Joined: 15 Aug 2005
Posts: 199
Location: New York
Followers: 1

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

 [#permalink]

Show Tags

New post 18 Jan 2006, 20:34
total number of groups if the three people are chosen randomly= 17C3=680

total number of groups if no professor is chosen = 10C3 = 120

total number of groups if at least a professor is chosen = 680-120 =560
1 KUDOS received
VP
VP
avatar
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 57 [1] , given: 0

 [#permalink]

Show Tags

New post 18 Jan 2006, 20:35
1
This post received
KUDOS
I think the answer is B 560.

You could do it in two ways:

Total combinations - Combinations without any professor = 17C3 - 10C3 =560

OR

Exactly One professor + Exactly 2 professors + Exactly 3 professors
= 7C1*10C2 + 7C2*10C1+ 7C3 = 315 + 210 + 35 = 560
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 382
Followers: 1

Kudos [?]: 69 [0], given: 0

 [#permalink]

Show Tags

New post 18 Jan 2006, 22:33
Option are

1 Prof + 2 Student C(7,1) * C(10, 2) +
2 Prof + 1 Student C(7,2) * C(10, 1) +
3 Prof C(7,3) +
CEO
CEO
User avatar
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 22

Kudos [?]: 205 [0], given: 0

 [#permalink]

Show Tags

New post 19 Jan 2006, 04:08
Atleast 1 prof = All - No Prof
= 17C3 - 10C3 = 680-120 = 560
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 29

Kudos [?]: 280 [0], given: 0

 [#permalink]

Show Tags

New post 19 Jan 2006, 09:45
Out of three places, 1 must be a professor. So let's consider the remaining two places.

Of those two places, we can either have: (professor, grad), (grad, grad) or (professor, professor)

Case 1: (professor, grad) -> Team will comprise 2 professors, 1 grad
# of ways = 7C2*10C1

Case 2: (grad, grad) -> Team will comprise 1 professor, 2 grads
# of ways = 7C1*10C2

Case 3: (professor, professor) -> Team will comprise 3 professors
# of ways = 10C3

Total # of ways = Case 1 + Case2 + Case 3
Manager
Manager
User avatar
Joined: 16 Feb 2007
Posts: 73
Followers: 1

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

Re: combination problem PS [#permalink]

Show Tags

New post 30 Aug 2007, 14:19
rlevochkin wrote:
A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

Plz, help me out with this, guys.

My solution: C(7;1)*C(16;2)=7*120=840
What's wrong with my reasoning here.


many posted the right solution here, which is 560. but could some permutation / combination expert please explain why the solution proposed above: C(7;1)*C(16;2)=7*120=840 is not applicable here?? i cant find an explanation...

thanks a lot

EDIT:

with the help of my little sister i found the flaw myself... :-) the problem is when you select 2 out of 16, i.e. C(16;2) then these two guys might be two professors or one professor and you cannot multiply them with 7 (the #professors) since only 5 or 6 professors would be left. therefore, as the correct solution suggest, you have differentiate between the scenarios 1prof/2studs, 2profs/1stud, 3profs.

Last edited by skkingdom on 30 Aug 2007, 14:49, edited 1 time in total.
Manager
Manager
avatar
Joined: 17 Apr 2007
Posts: 93
Followers: 1

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

Re: combination problem PS [#permalink]

Show Tags

New post 30 Aug 2007, 14:47
skkingdom wrote:
many posted the right solution here, which is 560. but could some permutation / combination expert please explain why the solution proposed above: C(7;1)*C(16;2)=7*120=840 is not applicable here?? i cant find an explanation...

thanks a lot


not so sure, but this how i understand it. you are trying to multiply the probabilities of choosing from the two overlaping sets. so you have many duplicates.
Manager
Manager
User avatar
Joined: 16 Feb 2007
Posts: 73
Followers: 1

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

Re: combination problem PS [#permalink]

Show Tags

New post 30 Aug 2007, 14:51
ankita wrote:
skkingdom wrote:
many posted the right solution here, which is 560. but could some permutation / combination expert please explain why the solution proposed above: C(7;1)*C(16;2)=7*120=840 is not applicable here?? i cant find an explanation...

thanks a lot


not so sure, but this how i understand it. you are trying to multiply the probabilities of choosing from the two overlaping sets. so you have many duplicates.


i think i just edited above at the same time you answered ;-) yes you are right, there is a considerable overlap, that makes the 840 possibilities incorrect
Re: combination problem PS   [#permalink] 30 Aug 2007, 14:51
Display posts from previous: Sort by

A three-person committee must be chosen from a group of 7

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.