FCP Rule:

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1st member 2nd Member 3rd Member

1st Member (at least 1 professor) = 4 ways
2nd Member = 9 ways
3rd Member = 8 ways!

Total ways = 4 x 9 x 8

or should i consider 3 possibilities of 1 Professor, 2 professors & 3 professors?

Can some one please help me solve this question via FCP rule?
Through combinations, I am able to solve.

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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

Explanation :

The problem with your FCP approach is it gives different combinations of the same committee :

COnsider the case where the committee consists of only 2 people, so according to your method, the answer would be 4 * 9 = 36.
What this implies is that with any Professor you will get 9 committees.
For eg with P1 : P1S1, P1S2, P1P2,... 9 such committees
with P2 : P2S1, P2S2, P2P1,... again 9 such committees.
What you find is that some of the committees are repeated twice such as P1P2, P2P1 and there is no point in counting any committee twice. To get around this problem, what you can do is break down the problem into three separate cases:

Case 1 : When only 1 professor is in the committee, the committee will consist of 1 professor and 2 students : 4 * 6* 5/2! = 120/2! = 60 (Dividing by 2! because each group of students is counted 2! times such as S1S2, S2S1,.... So dividing it ensures that every group is counted only once)

Case 2 : With 2 professors and 1 student :
4*3*6/2! = 72/2 = 36 (Every group of professor is counted 2! times)

Case 3 : all 3 professors :
4*3*2/3! = 24/6 = 4 (Every group of 3 professors is counted 3! times)

Adding all the 3 cases, we get : 60+36+4 = 100

madhusudhan237 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?

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

Good committees = total possible committees - bad committees.

Total possible committees:
From 10 people, the number of ways to choose 3 = 10C3 = (10*9*8)/(3*2*1) = 120.

Bad committees:
A bad committees does NOT include at least 1 professor.
In other words, a bad committee is composed solely of STUDENTS.
From the 6 students, the number of ways to choose 3 = 6C3 = (6*5*4)/(3*2*1) = 20.

Good committees = 120-20 = 100.

The correct answer is E.