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4 professors and 6 students are being considered for [#permalink]
27 Oct 2009, 05:28
1
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00:00
A
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Difficulty:
25% (medium)
Question Stats:
70% (02:12) correct
30% (01:13) wrong based on 406 sessions
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?
A. 36 B. 60 C. 72 D. 80 E. 100
So my line of thinking was the following:
4 x 9 x 8 = 288 (professor) (total remaining) (total remaining)
Obviously I'm wrong. Would anyone care to tell me, how my logic is flawed?
Re: 4 professors and 6 students are being considered for [#permalink]
23 May 2013, 06:29
2
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vishnuvardhan777 wrote:
In the Picture, Slot 1 could be filled in
4 C ways (read as four-c-one) 1
After one professor is selected then total remaining are 9 people (3 prof + 6 students) so slots 2 and 3 can be filled in
9 C * 2! ways 2
Please let me know, what is error in the logic
The red part is not correct. The correct formula is \((#Tot) C (#Slots)\), so in the case with 1 prof and 2 students we have: 4(#Prof) C 1(#SlotsP) * 6(#Stud) C 2(#SlotsS)
the committee must include one professor 1 professor, 2 students: \(4C1*6C2=60\) 2 professors, 1 student: \(4C2*6C1=36\) 3 professors, 0 students: \(4C3*6C0=4\)
Total number = 100
Hope it's clear _________________
It is beyond a doubt that all our knowledge that begins with experience.
Re: Why am I wrong? [#permalink]
27 Oct 2009, 12:14
1
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Expert's post
1
This post was BOOKMARKED
andershv wrote:
yuskay wrote:
andershv wrote:
My question is why the logic behind my proposed solution is wrong. Can anybody please explain?
you count up combinations twice.
4*9*8... and let me name 6 with students A,B,C,..,F after choosing a professor, when you choose Student A and B, you can choose A first, then B. Or B first, then A. Both combination are same.
I don't think that's the only mistake.
100 is a factor of = 2^2*5^2 288 is a factor of = 2^5*3^2.
These are very different numbers and you cannot divide 288 with something to get to 100.
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? A. 36 B. 60 C. 72 D. 80 E. 100
At least one professor out of 3: \(C^3_{10}\) (total # of selection of 3 out of 10) minus 6\(C^3_6\) (# of selection of 3 person from 6 students, which means zero professor) = \(120-20=100\);
OR: \(C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100\);
Answer: E.
But you already know this. Your question is different:
And, I think I understand what you mean: saying that "288 and 100 are very different numbers", you may imply that we can not divide 288 by any factorial to compensate duplications, as we almost always do when order doesn't matter and when we need to get rid of the same selections.
But not this time: because here we may have THREE DIFFERENT cases (1p2s; 2p1s or 3p0s), with different factorial correction in each. For example case when we have 1 professor and 2 students needs the different factorial correction than case when we have 2 professors and 1 student. Hence, you can not divide 288 by one factorial (say 2! or 3!) to fix it. So 288 has duplications but it can not be compensated just by dividing it by any factorial.
Re: Combos where did i go wrong4 professors and 6 students are b [#permalink]
29 Sep 2010, 16:08
1
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Jinglander wrote:
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?
Answer is 100 i got it wrong
this is what I did
choose one prof first 4 way to do this. Then choose 2 from the remain 9 so 2C9 thus 4*2C9
I could also see doing this the way that generates the right answer which is 3C10 - 6C3. They both seem right
Re: Why am I wrong? [#permalink]
19 Nov 2011, 08:33
1
This post received KUDOS
Here's quoting Ian Stewart.
Quote:
Say your professors are A, B, C and D, and we choose a committee of three professors. If we choose A first, then choose B and C, we get the same committee as if we choose B first, then choose A and C; the order of the three professors does not matter. In your solution, you are assuming that the order of the professors does, at least partly, matter - you're picking one professor as the 'first professor', and then you're selecting the rest of the committee. Because of that, you're overcounting.
The answer you give would be correct if the question assigned a position to one of the committee members - for example, if it asked 'If a three person committee will be chosen, and one professor must be chosen as the chairperson of the committee, how many committees can be chosen?" We're then assigning a position to one professor, and your solution would be correct.
Re: Why am I wrong? [#permalink]
15 Jan 2012, 12:24
1
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Expert's post
jgonza8 wrote:
Can someone confirm if my logic makes any sense? (or if I am making stuff up because I found the correct answer somewhere else?). So...here we go.
10!/7!3! - This one highlights that there will be a committee to be formed out of a pool of 10 people. 7 of which will not get in and 3 of which will get in. The total for this item is 120. However, the 120 does not take into consideration the constraint highlighted in the problem which says that the committee must include one professor. (In the denominator, 3!, accounts for 3! individuals...but we don't know which ones are professors or students).
In order to accommodate for the constraint, we have to go out and calculate the composition of the committee if it didn't include professors. So we come up with...
6!/3!3! - 6 possible students that can get in. 3 which make the cut and the other 3 which do not. The total for this calculation yields 20.
So...now that we know that 20 only considers students, we need to go back and subtract that from the original calculation since that is a constraint in the problem. (i.e. the problem says it must include at least one professor
Based on that 120-20 = 100...(E).
Thoughts?
All is correct except the red part. The committee has to include at least 1 professor (not one as you've written).
{The committees with at least one professor} = {Total committees possible} - {The committees with zero professors} (so minus the committees with only students in them).
{The committees with at least one professor} = {The committees with 1 professor / 2 students} + {The committees with 2 professors / 1 student} + {The committees with 3 professors / 0 students}:
Re: 4 professors and 6 students are being considered for [#permalink]
08 Sep 2013, 22:53
1
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Expert's post
tyagigar wrote:
why i can not do like this:
formation has to have at-least 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.
Re: 4 professors and 6 students are being considered for [#permalink]
24 Oct 2015, 16:34
1
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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!
Re: 4 professors and 6 students are being considered for [#permalink]
25 Oct 2015, 11:13
1
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Expert's post
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.
Re: Why am I wrong? [#permalink]
27 Oct 2009, 07:03
andershv wrote:
My question is why the logic behind my proposed solution is wrong. Can anybody please explain?
you count up combinations twice.
4*9*8... and let me name 6 with students A,B,C,..,F after choosing a professor, when you choose Student A and B, you can choose A first, then B. Or B first, then A. Both combination are same.
Re: Why am I wrong? [#permalink]
27 Oct 2009, 07:20
yuskay wrote:
andershv wrote:
My question is why the logic behind my proposed solution is wrong. Can anybody please explain?
you count up combinations twice.
4*9*8... and let me name 6 with students A,B,C,..,F after choosing a professor, when you choose Student A and B, you can choose A first, then B. Or B first, then A. Both combination are same.
I don't think that's the only mistake.
100 is a factor of = 2^2*5^2 288 is a factor of = 2^5*3^2.
These are very different numbers and you cannot divide 288 with something to get to 100.
Re: Combos where did i go wrong4 professors and 6 students are b [#permalink]
29 Sep 2010, 16:21
Jinglander wrote:
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?
Answer is 100 i got it wrong
this is what I did
choose one prof first 4 way to do this. [highlight]Then choose 2 from the remain 9 so 2C9 thus 4*2C9[/highlight]
I could also see doing this the way that generates the right answer which is 3C10 - 6C3. They both seem right
The above highlighted portion is incorrect. You should choose 2 from the 6 students and not the remaining 9 [3 Prof and 6 Students] since the 9 could introduce few more professors.
"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?"
What kind of problem is this? A probability/possibility problem? I superficially understand the answer but I don't understand the concept deeply:
10!/(7!3!) - 6!/(3!3!) = 100
Any explanation or being pointed to the right direction in the Manhattan GMAT books would be appreciated.
Re: Why am I wrong? [#permalink]
15 Jan 2012, 11:07
Can someone confirm if my logic makes any sense? (or if I am making stuff up because I found the correct answer somewhere else?). So...here we go.
10!/7!3! - This one highlights that there will be a committee to be formed out of a pool of 10 people. 7 of which will not get in and 3 of which will get in. The total for this item is 120. However, the 120 does not take into consideration the constraint highlighted in the problem which says that the committee must include one professor. (In the denominator, 3!, accounts for 3! individuals...but we don't know which ones are professors or students).
In order to accommodate for the constraint, we have to go out and calculate the composition of the committee if it didn't include professors. So we come up with...
6!/3!3! - 6 possible students that can get in. 3 which make the cut and the other 3 which do not. The total for this calculation yields 20.
So...now that we know that 20 only considers students, we need to go back and subtract that from the original calculation since that is a constraint in the problem. (i.e. the problem says it must include at least one professor
Re: 4 professors and 6 students are being considered for [#permalink]
23 May 2013, 07:10
vishnuvardhan777 wrote:
Thanks Zarrolou; I understood the explanation.
Can you please reason, why i should not take the second part as 9c2 (read as nine see two).
(I am considering the case 1 Prof and 2 Students, the reasoning is the same for the others)
You have \(3\) slots. \(1\) will be occupied by professor and \(2\) by a students. Lets focus on the professors: there are \(4\) of those and you have \(1\) slot => tot comb = \(4C1\). Same thing for the students: there are \(6\) of those and you have \(2\) slots => tot comb = \(6C2\).
Tot cases of 1 Prof and 2 Stud = \(4C1*6C2=60\)
In your method you correctly choose 1 prof in \(4C1\) ways, but you cannot take \(9C2\) as the second term because you are not choosing from a pool of 9, but from a pool of 6 (students ONLY in this case). So the correct form is \(6C2\)
Hope it makes sense, let me know _________________
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