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

 It is currently 15 Oct 2019, 20:15

### 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

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

# A university needs to select a nine-member committee on extr

Author Message
TAGS:

### Hide Tags

Intern
Joined: 21 Jan 2009
Posts: 4
A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

08 Jun 2010, 13:30
1
14
00:00

Difficulty:

35% (medium)

Question Stats:

73% (02:10) correct 27% (02:39) wrong based on 386 sessions

### HideShow timer Statistics

A university needs to select a nine-member committee on extracurricular life, whose members must belong either to the student government or to the student advisory board. If the student government consists of 10 members , the student advisory board consists of 8 members and 6 students hold membership in both organizations , how many different committee are possible.

A. 72
B. 110
C. 220
D. 720
E. 1096
Math Expert
Joined: 02 Sep 2009
Posts: 58340

### Show Tags

08 Jun 2010, 14:04
3
4
sumana wrote:
A university needs to select a nine-member committe on extracurricular life, whose members must belong either to the student government or to the student advisory board. If the student government consists of 10 members , the student advisory board consists of 8 members and 6 students hold membership in both organiztions , how many different committes are possible.

a) 72
b) 110
c) 220
d) 720
e) 1096

Two groups:
Student government = 10;
Both = 6.

Total # of members of either of group is: 10+8-6=12.

# of ways to choose 9 people out of 12 is $$C^9_{12}=\frac{12!}{9!3!}=220$$.

Hope it helps.
_________________
##### General Discussion
Manager
Joined: 25 Sep 2012
Posts: 232
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31
GMAT 2: 680 Q48 V34
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

15 Feb 2014, 07:50
is the difficulty really sub600?
I find it more difficult.
Director
Joined: 03 Feb 2013
Posts: 834
Location: India
Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

15 Feb 2014, 10:30
b2bt wrote:
is the difficulty really sub600?
I find it more difficult.

I think it is 600-700 level. Compared to OG which are sub 600, this question is actually good.

Combination of two concepts (Venn Diagram and Combination) -> 600-700 level.
_________________
Thanks,
Kinjal

My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961

Intern
Joined: 14 Oct 2013
Posts: 43
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

25 Apr 2015, 11:20
How is 9C12 = 220? I keep getting just 20.
Director
Joined: 03 Feb 2013
Posts: 834
Location: India
Concentration: Operations, Strategy
GMAT 1: 760 Q49 V44
GPA: 3.88
WE: Engineering (Computer Software)
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

25 Apr 2015, 11:35
1
healthjunkie wrote:
How is 9C12 = 220? I keep getting just 20.

12C9 = 12 ! /(9! * 3!) = 10 * 11 * 12/ 2*3 = 10* 11 * 2 = 220.

I hope this is clear.
_________________
Thanks,
Kinjal

My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961

Intern
Joined: 01 Apr 2015
Posts: 10
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

26 Apr 2015, 13:00
Hello,

Step 1:
student government consists of 10 members + the student advisory board consists of 8 members (so add them): 10 + 8 = 18

Step 2:
6 students hold membership in both organizations ( so subtract from total in step 1): 18- 6 = 12

Step 3:
How many different committee are possible if 9 will be selected.
So 12! / 9! 3! (9 will be selected and 3 will not be selected).

12 x 11 x 10 / 3 x 2 x 1 = 220 Answer is C.
Intern
Joined: 19 Oct 2013
Posts: 7
Location: India
Concentration: Finance, Marketing
GMAT 1: 720 Q50 V37
WE: Analyst (Internet and New Media)
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

17 Jul 2016, 19:07
Total number of students = 10+8-6 = 12. Ways to select 9 out of 12 = 12C9 = 12*11*10/3*2=220

Sent from my ONE A2003 using Tapatalk
Manager
Joined: 17 Jun 2015
Posts: 196
GMAT 1: 540 Q39 V26
GMAT 2: 680 Q46 V37
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

12 Sep 2016, 11:50
No. of students in government = 10
No. of students in advisory board = 8
No. of students in both = 6
Total = 10 + 8 - 6 = 12

Total number of 9 member committees would be 9C3 = 220.

I wonder if the question could be made a 700 difficulty one by adding constraints. Such as exactly one student should be from the one of the two groups. Food for thought!
_________________
Fais de ta vie un rêve et d'un rêve une réalité
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2817
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

23 Jan 2018, 09:01
sumana wrote:
A university needs to select a nine-member committee on extracurricular life, whose members must belong either to the student government or to the student advisory board. If the student government consists of 10 members , the student advisory board consists of 8 members and 6 students hold membership in both organizations , how many different committee are possible.

A. 72
B. 110
C. 220
D. 720
E. 1096

The number of students who belong either to the student government or to the student advisory board or both is:

Total = # Government + # Advisory - # Both + # Neither

10 + 8 - 6 + 0 = 12 students.

So the number of ways to select the 9-member committee out of 12 possible candidates is 12C9:

12!/[(12-9)! x 9!]

12!/(3! x 9!) = (12 x 11 x 10)/3! = (12 x 11 x 10)/(3 x 2) = 2 x 11 x 10 = 220

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Joined: 07 Feb 2017
Posts: 7
Location: India
GPA: 4
WE: Sales (Manufacturing)
Re: A university needs to select a nine-member committee on extr  [#permalink]

### Show Tags

28 Jan 2018, 11:56
Kudos pls, if you like the post
Attachments

File comment: Table format was my approach.
Hope this is one other way of solving it

Soln-Fighter.jpg [ 1.09 MiB | Viewed 807 times ]

Re: A university needs to select a nine-member committee on extr   [#permalink] 28 Jan 2018, 11:56
Display posts from previous: Sort by