4 professors and 6 students are being considered for : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 14:33

### 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

# 4 professors and 6 students are being considered for

Author Message
Director
Joined: 10 Feb 2006
Posts: 658
Followers: 3

Kudos [?]: 459 [0], given: 0

4 professors and 6 students are being considered for [#permalink]

### Show Tags

06 Jun 2007, 20:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committee be formed if it has to include at least 1 professor ?
_________________
GMAT the final frontie!!!.
_________________

GMAT the final frontie!!!.

Manager
Joined: 17 Oct 2006
Posts: 52
Followers: 0

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

### Show Tags

06 Jun 2007, 21:02
Since a committee containing prof. A& B is similar to committee containing prof. B&A so it means that order doesn't matter and therefore it is a combinations problem.
Now, this question has three possible committe compositions, 1) with 1 prof and 2 studets, 2) with 2 prof and 1 students and 3) with 3 prof. and no students in the committee.
SO working the maths now
1 prof from 4 can be chosen in 4C1 wasy and 2 students from 6 studnets can be chosen in 6C2 ways. Similarly setting up maths for other two combinations, we get
(4C1*6C2)+(4C2*6C1)+(4C3)
=(4!/3!*6!/2!*4!)+(4!/2!*2!*6!/5!)+(4!/3!)
=(4*15)+(6*6)+(4)
=60+36+4
=100 Ans
Ciao
06 Jun 2007, 21:02
Display posts from previous: Sort by