M25 #3 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 19:42

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M25 #3

Author Message
Manager
Affiliations: Beta Gamma Sigma
Joined: 14 Aug 2008
Posts: 210
Schools: Harvard, Penn, Maryland
Followers: 6

Kudos [?]: 65 [0], given: 3

### Show Tags

23 May 2009, 21:33
not that I don't agree with the answer, but could someone please explain the formulas, I see that it may help me in the future to know them, and I dont have a stats textbook handy..

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?
* 36
* 60
* 72
* 80
* 100

First, consider an unconstrained version of the question: how many committees of 3 are possible? The answer is $$C_{10}^3 = \frac{10!}{(7!3!)} = 120$$ . Now subtract the number of committees that consist entirely of students i.e. $$C_{6}^3 = \frac{6!}{(3!3!)} = 20$$ . The final answer is $$C_{10}^3 - C_6^3 = 120 - 20 = 100$$ .
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14457
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3726

Kudos [?]: 23036 [0], given: 4515

### Show Tags

24 May 2009, 22:47

Combinations: http://en.wikipedia.org/wiki/Combination

Permutations: http://en.wikipedia.org/wiki/Permutation

Both of these come from the Combinatorics field of Math and answer a specific question of how many combinations are possible in any given situation based on the total number of members and any limitations.

The main difference between C and P is that in Combinations order does not matter and Permutations order matters - see this article for the easy explanation of that difference: http://betterexplained.com/articles/eas ... binations/
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

GMAT Club Premium Membership - big benefits and savings

Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14457
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3726

Kudos [?]: 23036 [0], given: 4515

### Show Tags

24 May 2009, 22:55

Continue discussions in the appropriate thread.
_________________

Founder of GMAT Club

US News Rankings progression - last 10 years in a snapshot - New!
Just starting out with GMAT? Start here...
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

GMAT Club Premium Membership - big benefits and savings

Re: M25 #3   [#permalink] 24 May 2009, 22:55
Similar topics Replies Last post
Similar
Topics:
1/(2 - sqrt(3)) = ? (m25#36) 2 15 Feb 2012, 06:04
19 m25 q.3 21 21 Feb 2009, 15:34
18 m25#02 29 17 Oct 2008, 13:28
27 m25 #22 21 20 Oct 2008, 14:27
17 M25 #29 23 09 Oct 2008, 06:34
Display posts from previous: Sort by

# M25 #3

Moderator: Bunuel

 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®.