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# 4 professors and 6 students are being considered for

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Manager
Joined: 08 Oct 2009
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4 professors and 6 students are being considered for  [#permalink]

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27 Oct 2009, 06:28
3
7
00:00

Difficulty:

35% (medium)

Question Stats:

73% (01:45) correct 27% (02:00) wrong based on 630 sessions

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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?

Thanks!
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Posts: 49968
Re: Why am I wrong?  [#permalink]

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27 Oct 2009, 13:14
2
3
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: $$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100$$;

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.

Hope it's clear.
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Re: Why am I wrong?  [#permalink]

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27 Oct 2009, 06:43
6
1
the way I solved it is:

the committee should have atleast 1 professor, so possible committee combinations can include

1 professor 2 students or
2 professors 1 student or
3 professors 0 students

and we can select above combinations as

4C1 * 6C2 + 4C2*6C1 +4C3 = 60 + 36 +4 = 100

I am not sure how to explain why your approach is wrong 4 * 9 * 8
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Re: Why am I wrong?  [#permalink]

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27 Oct 2009, 07:37
3
1
It should be easier to solve "at least" questions in this way.

All combinations are 10C3,

Combinations of all students are 6C3,

10C3-6C3=at least one professor combi.

thanx
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Joined: 21 Oct 2009
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Re: Why am I wrong?  [#permalink]

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27 Oct 2009, 08:03
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.
Manager
Joined: 08 Oct 2009
Posts: 67
Location: Denmark, Europe
Schools: Darden Class of 2012
Re: Why am I wrong?  [#permalink]

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27 Oct 2009, 08:20
1
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.
Manager
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Re: Why am I wrong?  [#permalink]

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27 Oct 2009, 15:31
Yes, it's clear now!

Thanks Bunuel, you're a great sport!
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Re: Combos where did i go wrong4 professors and 6 students are b  [#permalink]

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29 Sep 2010, 17:08
1
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

combinations:

1 professor 2 students

students: 6 x 5 / 2 = 15
15*4 = 60

2 professors 1 student

4 x 3 / 2 = 6

6*6 = 36

3 professors
4/1 = 4

60 + 36 + 4 = 100
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Re: Combos where did i go wrong4 professors and 6 students are b  [#permalink]

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29 Sep 2010, 17:21
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.

Answer is $$4C1*6C2 + 4C2*6C1 + 4C3 => 100$$ ways
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Posts: 49968
Re: Need Help - not sure what kind of problem this is  [#permalink]

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04 Feb 2011, 07:06
Merging similar topics. Please refer to the solutions above and ask if anything remains unclear.

Provide answer choices for PS questions.

randyrosales wrote:
From the GMAT Club Test, m25

"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.

As for you follow up question: it's a combinatorics problem. Check Combinatorics chapter of Math Book (gmat-math-book-87417.html) by Walker for more: math-combinatorics-87345.html

PS questions on combinations: search.php?search_id=tag&tag_id=31
DS questions on combinations: search.php?search_id=tag&tag_id=52

Hope it helps.
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Re: Why am I wrong?  [#permalink]

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19 Nov 2011, 09:33
1
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.

The mistake is clear now.
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Re: Why am I wrong?  [#permalink]

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15 Jan 2012, 12:07
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

Based on that 120-20 = 100...(E).

Thoughts?
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Re: Why am I wrong?  [#permalink]

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15 Jan 2012, 13:24
1
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).

$$C^3_{10}-C^3_6=\frac{10}{7!*3!}-\frac{6!}{3!*3!}=120-20=100$$.

Or direct approach:

{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}:

$$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100$$.

Hope it's clear.
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Re: 4 professors and 6 students are being considered for  [#permalink]

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23 May 2013, 07:02
I solved this in the below fashion

Attachment:

GmatQQ.png [ 1.32 KiB | Viewed 13755 times ]

In the Picture, Slot 1 could be filled in

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

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23 May 2013, 07:29
2
vishnuvardhan777 wrote:

In the Picture, Slot 1 could be filled in

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

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23 May 2013, 08:01
Thanks Zarrolou; I understood the explanation.

Can you please give me the reason: why i should consider second part as 9c2 (read as nine see two).
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Re: 4 professors and 6 students are being considered for  [#permalink]

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23 May 2013, 08:10
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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Re: Why am I wrong?  [#permalink]

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11 Jun 2013, 15:12
Sorry to muddle the thread with simple maths:

I'm trying to follow the working in $$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100$$

If $$C^n_r$$ is $$\frac{r!}{n!(r-n)!}$$ then,

$$C^1_4 = 4$$
$$C^2_6+C^2_4 = 15 + 6$$
$$C^1_6+C^3_4 = 6 + 4$$

Sum = 35

What am I doing wrong?

I found this question a fair bit harder than the examples published by GMAC (in books and software). Is it more indicative of the exam?
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Re: Why am I wrong?  [#permalink]

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11 Jun 2013, 16:45
stormbind wrote:
Sorry to muddle the thread with simple maths:

I'm trying to follow the working in $$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=100$$

If $$C^n_r$$ is $$\frac{r!}{n!(r-n)!}$$ then,

$$C^1_4 = 4$$
$$C^2_6+C^2_4 = 15 + 6$$
$$C^1_6+C^3_4 = 6 + 4$$

Sum = 35

What am I doing wrong?

I found this question a fair bit harder than the examples published by GMAC (in books and software). Is it more indicative of the exam?

You missed multiplication signs: $$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=4*15+6*6+4=60+36+4=100$$.

As for your other question, you can get an easy as well as hard (like this one) combination questions on the test.
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Re: 4 professors and 6 students are being considered for  [#permalink]

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08 Sep 2013, 00:50
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
Re: 4 professors and 6 students are being considered for &nbs [#permalink] 08 Sep 2013, 00:50

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