Jun 29 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Jun 30 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Jul 01 08:00 AM PDT  09:00 AM PDT Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Jul 01 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55801

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
08 Sep 2013, 23:53
tyagigar wrote: why i can not do like this:
formation has to have atleast 1 professor : we chose no of ways i can select 1 professor : 4c1= 4 ways no we have to select 2 more people and since our requirement of atleast 1 professor is already met can we not select the remaining 2 people out of 9 in 9c2 ways i.e (9*8*7!)/(2!*7!)=36 ways so total no of ways=4*36
i know i am doing something wrong , can i get some help please The number you get will have duplications. Consider the group of three professors {ABC} Say you select professor A with 4C1, next you can get professors B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group. Does this make sense?
_________________



Intern
Joined: 11 Jul 2013
Posts: 31

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
09 Sep 2013, 07:35
Bunuel wrote: tyagigar wrote: why i can not do like this:
formation has to have atleast 1 professor : we chose no of ways i can select 1 professor : 4c1= 4 ways no we have to select 2 more people and since our requirement of atleast 1 professor is already met can we not select the remaining 2 people out of 9 in 9c2 ways i.e (9*8*7!)/(2!*7!)=36 ways so total no of ways=4*36
i know i am doing something wrong , can i get some help please The number you get will have duplications. Consider the group of three professors {ABC} Say you select professor A with 4C1, next you can get professors B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group. Does this make sense? thanks a lot, this makes perfect sense



Intern
Joined: 03 Nov 2013
Posts: 1

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
22 Apr 2014, 03:47
Can any one tell where the below approach went wrong?
Selecting one professor from 4 * select any two from the remaining nine members 4c1 * 9c2 = 4 * 9*8/2*1 = 144



Math Expert
Joined: 02 Sep 2009
Posts: 55801

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
22 Apr 2014, 03:53
saiyanVegeta wrote: Can any one tell where the below approach went wrong?
Selecting one professor from 4 * select any two from the remaining nine members 4c1 * 9c2 = 4 * 9*8/2*1 = 144 Check here: 4professorsand6studentsarebeingconsideredfor8586520.html#p1265096Hope it helps.
_________________



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 410
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
25 Feb 2015, 11:28
In this problem, may I ask why we divided by the part in red?
10!/(7!3!)  6!/(3!3!) = 100
I understand the idea, and the first part is perfectly clear. I understand subtracting the possibility of no teachers, but what does 3!3! mean? Now, I understand why it is not 6!/3!, but in this case 3!3! seems to mean 3 students selected and 3 not selected. Or not...?



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14394
Location: United States (CA)

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
25 Feb 2015, 17:11
Hi pacifist85, The approach that you are referencing is meant to do the entire calculation in one gigantic "move", which I'm not a big fan of. It's simple enough to break the calculation into 3 "pieces" and then add together the results. However, since you asked though, this calculation is based on two ideas: 1) You have to choose 3 people from a group of 10 (that's the first part of the calculation) 2) Since the group has to include at least 1 professor, you CAN'T have a group that is made up of 3 students. Since there are 6 total students, there are 6!/3!3! different groups of 3 students that could be formed. We have to REMOVE those options from the prior part of the calculation. This gives us: (All possible groups of 3)  (All groups of 3 that are JUST students) = (groups of 3 with at least 1 professor). GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/



Intern
Joined: 15 Feb 2015
Posts: 13

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
27 Feb 2015, 12:30
all possibitlity: C(10 3) = 120 possibility for no professor: C(6 3) = 20 possibility for at least one professor = 120  20 1= 100



Intern
Joined: 05 Aug 2015
Posts: 40

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
24 Oct 2015, 16:46
Can someone please help me understand why is 6C3 (choose 3 students from 6 students) = 6!/(3!3!) so 6*5*4/(3*2) >20 ways, instead of simply 6*5*4 following the reasoning that choice #1  you can pick from 6 students, choice #2  you can pick from 5 students, and finally choice #3  you can pick from 4 students. Thanks so much!
_________________
Working towards 25 Kudos for the Gmatclub Exams  help meee I'm poooor



Intern
Joined: 27 Jun 2015
Posts: 6
Concentration: Marketing, Entrepreneurship

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
24 Oct 2015, 17:34
happyface101 wrote: Can someone please help me understand why is 6C3 (choose 3 students from 6 students) = 6!/(3!3!) so 6*5*4/(3*2) >20 ways, instead of simply 6*5*4 following the reasoning that choice #1  you can pick from 6 students, choice #2  you can pick from 5 students, and finally choice #3  you can pick from 4 students.
Thanks so much! Lets use a simplier format first to help with the reasoning: How many ways to choose 3 students from 3 students? Based on your approach above we have: 1. 3x2x1/(3x2x1) = 1 or 2. 3x2x1 = 6 Choice (2) includes all the repeats and this is the answer if order matters meaning, abc is different from bca Choice (1) includes the option where order does not matter meaning abc is the same as bca Back to the original question: Since we are asked to form a group, order doesn't matter so we can ignore all the repeats: A group with Bill, Elon and Tim is the same group with Elon, Tim and Bill!



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14394
Location: United States (CA)

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
25 Oct 2015, 12:13
Hi happyface101, In these types of questions, it's important to determine whether the 'order' of the people matters. Often, the prompt will use certain vocabulary when referencing a permutation ("arrangements") or combination ('groups"). If you aren't given any obvious 'clues' to work with, you have to think critically about what the question is actually asking you to do. When forming a basic 'group', the order of the group won't matter: eg A group made up of A, B and C is the SAME group as one that is made up of B, C and A. Thus, we cannot count this group more than once. IF... we're assigning roles to the members of a group, then the order DOES matter. eg. We have a group of 3 people; one is the president, one is the secretary and one is the treasurer. In this scenario, the order matters, so there are (3)(2)(1) = 6 groups. Interestingly enough, many of these types of prompts can be solved with EITHER approach, but you have to adjust your calculations accordingly. GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/



Intern
Joined: 28 Aug 2015
Posts: 18
Location: Norway
WE: Information Technology (Consulting)

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
02 Nov 2015, 12:54
happyface101 wrote: Can someone please help me understand why is 6C3 (choose 3 students from 6 students) = 6!/(3!3!) so 6*5*4/(3*2) >20 ways, instead of simply 6*5*4 following the reasoning that choice #1  you can pick from 6 students, choice #2  you can pick from 5 students, and finally choice #3  you can pick from 4 students.
Thanks so much! Hi, See the Question, AT LEAST 1 Professor In every scenario, you need to count at least 1 or more than 1 professor, to obtain 3 Supervisory committee member. Case 1. : 1 Professor, 1 student, 1 Student = 3 Total member. Now, using the slot method or directly using combination formula, you can select 1 Professor out of 4 {4c1}, in 4 way. Similarly you can select 2 student out of 6 by {6c2} = 15 way . Now, as per Multiplication Principle, 15 * 4 = 60. Case 2. : 2 Professor, 1 student = 3 total member. Using the Combination formula, 2 Professor out of 4 {4c2}, in 6 way. Similarly for 1 student out of 6 by {6c1} = 6 way . > 6 * 6 = 36. Case 3: Similarly we can go for 3 professor (remember at least one and no limitation till three, as only 3 Supervisory committee member required), without selecting any student. This can done in {6c3} = 4 way. Add, case 1 + Case 2 + case3 > 60 + 36 +4 = 100, is the correct answer.
_________________
Thanks Sumit kumar If you like my post Kudos Please



Current Student
Joined: 02 Jun 2015
Posts: 77
Location: Brazil
Concentration: Entrepreneurship, General Management
GPA: 3.3

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
02 Nov 2015, 15:22
I did a little bit different.
Total number of ways: C = 10! / 3! 7! = 120
Total number of ways that I cannot have any professor, just students: 6/10 * 5/9 * 4/8 = 1/6
1/6*120 = 20 ways that I will only have students
Total number  only students = At least one professor 120  20 = 100 ways



NonHuman User
Joined: 09 Sep 2013
Posts: 11441

Re: 4 professors and 6 students are being considered for
[#permalink]
Show Tags
30 Jan 2019, 10:06
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: 4 professors and 6 students are being considered for
[#permalink]
30 Jan 2019, 10:06



Go to page
Previous
1 2
[ 33 posts ]



