M25-03

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

Bunuel · Accepted Answer

Official Solution:

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

bennettthecat · Answer

I still don't understand - this notation is not something I'm familiar with. Is there another way to explain?

pocketvenus · Answer

Hi,

I thought about the answer in terms of MGMAT's Combinatorics and Probability explanation. This is similar to Bunuel's explanation, but without the notation.

First you need to determine the number of ways that the members can be chosen. The total # of people being considered is 10. That's 10 choices that can be made, so 10!.

From these 10 people, 3 people must be chosen, which also means that 7 people are not chosen. Therefore the equation is written as:

\frac{10!}{3! * 7!}

The above is if everyone on in the group were the same and we were only choosing 3 people to be on the supervisory committee. However, we know that both students and professors are being considered. And more specifically, at least 1 professor must be chosen. This also implies that it could be 1 professor or 2 or even 3. This makes the problem complicated because we have to determine the number of arrangements for all 3 possibilities. An easier way is to determine the number of arrangements if there were no professors chosen. (This is in the same vein as P  = 1 - P ).

There are 6 students, so again, 6! arrangements. 3 can be on the committee and 3 cannot. So,  \frac{6!}{3! * 3!} . This is actually a possibility that we do not want (the prompt asks for at least 1 professor, this gives us the # of arrangements for 0 professors). So we want to subtract this:

\frac{10!}{3!7!} - \frac{6!}{3! 3!}  = 100.

Hope this helpful!

ronash · Answer

Hi Bunuel,
What is wrong with the following:
First select 1 professor and then for the remaining 2 position select any from the 9(6+3) left.
4C1 * 9C2

amanvermagmat · Answer

Hi

In that case there will be repetitions. Say professors are A, B, C, D and students are p, q, r, s, t, u.

When you fix one professor as A, and select 2 out of remaining (9C2) - this will give you 36 cases, and these 36 cases will include such as ABC, ABD, ACD, ABp, ACq,.. etc

Now when you fix one professor as B, and select 2 out of remaining (9C2) - this will again give you 36 cases, including BAC, BAD, BAp... etc

We can clearly see that there are some repetitions in first and second case; and similarly there will be repetitions in further cases. To avoid this overlap, that method by Bunuel is simple and best I think. (there are other ways too but probably more complicated than that method)

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

Gg153 · Answer

Bunuel 
 
Why do we not divide each combination by  2!  as we did  here ?

Bunuel · Answer

Good question. The reason we do not divide by 2! in the committee problem is because we are selecting a  single  group of 3 people, not dividing people into multiple groups. The order of selection does not matter, and there is only one group being formed.

In the other problem with 6 boys, we are dividing them into two separate teams of 3. Since the teams are indistinguishable, selecting Team A as {ABC} and Team B as {DEF} is the same as selecting Team A as {DEF} and Team B as {ABC}. To avoid counting such duplicates, we divide by 2!.

In short:

Committee problem: selecting one group, no need to divide by 2! 
 Team division: creating two indistinct groups, so divide by 2! to avoid repeats.

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