combination or permutation

Question

It happens that a graduate student in theoretical/mathematical physics is looking for five members of his dissertation committee. He has been working closely with three professors in the mathematics department, and 5 professors in the physics department on his dissertation research.

For comic relief, can you figure out how may ways can this hapless graduate student can choose among his beloved professors if the chair of the committee must be a mathematician, and the rest of the committee can be a mix of mathematicians and physicists?

Answer 1

Let the total set of possible people be:  M1, M2, M3, P1, P2, P3, P4, P5

The chair must come from the Math Department:  There are three choices.

The other four come from the remaining 7 (in any permutation; it’s a combination), so there are

      C(7,4) combinations

      C(n,r) = n! / (r! * (n-r)!)   =   n(n-1)(n-2). . . (n – r + 1) / r!

      C(7,4) = 7! / (4! * 3!)    = ( 7*6*5*4) / (4*3*2*1)   =  35

So, for each of the 3 choices for the chair, there are 35 choices for the rest of the committee.  That gives 3*35 = 105 total combinations.

Of course, a smart graduate student in theoretical/mathematical physics will realize the value of individuals and of interactions between 2, 3, 4 or even 5 people and be able to assign weights (values) to how such a group will interact in evaluating him and his dissertation (most professors have “soap boxes” and “hot buttons” and “favorite students.”).

Answer 2

3*7!/(4!*3!)=105

 

The head can be chosen in 3 ways.  Then choose 4 out of 7.  So 3*C(7,4).