From a group of 4 professors and 6 students, a supervisory committee

Question

From a group of 4 professors and 6 students, a supervisory committee of 3 members is to be assembled. If the committee must include at least one professor, how many ways can the 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!

srini123 · Accepted Answer

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

Bunuel · Answer

From a group of 4 professors and 6 students, a supervisory committee of 3 members is to be assembled. If the committee must include at least one professor, how many ways can the committee be formed?

A. 36
B. 60
C. 72
D. 80
E. 100

{The number of committees with at least one professor} = 
 
= {The total number of possible committees} minus {The number of committees with no professors} (i.e., committees made up solely of students).
 
Using this approach, calculate  C^3_{10}  (the total ways to select 3 out of 10) and subtract  C^3_6  (the ways to choose 3 individuals exclusively from the 6 students, hence no professors):
 
 C^3_{10}-C^3_6=\frac{10!}{7!*3!}-\frac{6!}{3!*3!}=120-20=100 .
 
Alternatively, we can use a direct method:
 
{The number of committees with at least one professor} = 
 
= {The committees with 1 professor and 2 students} plus {The committees with 2 professors and 1 student} plus {The committees with 3 professors and no students}:
 
This gives:  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: when saying that "288 and 100 are very different numbers," you may imply that we cannot divide 288 by any factorial to compensate for duplications, as we almost always do when the 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 a different factorial correction in each. For example, the case when we have 1 professor and 2 students needs a different factorial correction than the case when we have 2 professors and 1 student. Hence, you cannot divide 288 by one factorial (say 2! or 3!) to fix it. So, 288 has duplications, but it cannot be compensated just by dividing it by any factorial.

Hope it's clear.

yuskay · Answer

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

yuskay · Answer

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.

Augustus · Answer

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.

Augustus · Answer

Yes, it's clear now!

Thanks Bunuel, you're a great sport!

nonameee · Answer

Here's quoting Ian Stewart.

The mistake is clear now.

jgonza8 · Answer

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?

Bunuel · Answer

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.

pacifist85 · Answer

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...?

EMPOWERgmatRichC · Answer

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

happyface101 · Answer

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!

Thanks so much!

EMPOWERgmatRichC · Answer

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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