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Re: 4 professors and 6 students are being considered for
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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?
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Re: 4 professors and 6 students are being considered for
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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



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Re: 4 professors and 6 students are being considered for
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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



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Re: 4 professors and 6 students are being considered for
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22 Apr 2014, 03:53



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Re: 4 professors and 6 students are being considered for
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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...?



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Re: 4 professors and 6 students are being considered for
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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
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Re: 4 professors and 6 students are being considered for
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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



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Re: 4 professors and 6 students are being considered for
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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!
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Re: 4 professors and 6 students are being considered for
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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!



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Re: 4 professors and 6 students are being considered for
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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
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Re: 4 professors and 6 students are being considered for
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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.
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Re: 4 professors and 6 students are being considered for
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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



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Re: 4 professors and 6 students are being considered for
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30 Jun 2017, 04:30
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