# Thread: Factorials - Special Functions Story Prob

1. ## Factorials - Special Functions Story Prob

I think it is the way this is worded.. but I just can't really understand what they are even asking me to figure out. It is under the functions and graphs section and I know it has to do with factorials. I would really appreciate any help anyone can offer on this problem.

The business mathematics class has elected a grievance committee of four to complain to the faculty about the introduction of factorial notation into the course. They decide that they will be more effective if they label themselves as members A,G,M, and S where member A will lobby faculty with surnames A through F, member G will lobby faculty members G through L, and so on. In how many ways can the committee so label its members? In how many ways can a committee of five label itself with five different labels.

2. Originally Posted by rizzod206
I think it is the way this is worded.. but I just can't really understand what they are even asking me to figure out. It is under the functions and graphs section and I know it has to do with factorials. I would really appreciate any help anyone can offer on this problem.

The business mathematics class has elected a grievance committee of four to complain to the faculty about the introduction of factorial notation into the course. They decide that they will be more effective if they label themselves as members A,G,M, and S where member A will lobby faculty with surnames A through F, member G will lobby faculty members G through L, and so on. In how many ways can the committee so label its members? In how many ways can a committee of five label itself with five different labels.
"Fundamental Principle of Counting": If you can do $A_1$ in $n_1$ ways, $A_2$ in $n_2$ ways, ..., $A_k$ in $n_k$ ways, then you can do all of them in $n_1n_2\cdot\cdot\cdot n_k$ ways.

You can assign the label "A" to any of the 5 members so you can do that in 5 ways. Once that has been done, there are 4 members not yet assigned so you can assign the label "G" in 4 ways, then assign the label "M" in 3 ways, then assign the label "S" in 2 ways: 5(4)(3)(2)= 5! ways. Of course, that leaves the 5th member without a label. This is also the binomial coefficient $\left( \begin{array}{c}5 \\ 4 \end{array} \right) = \frac{5!}{4!(5-4)!}$.

Exactly the same analysis with 5 labels leads to exactly the same answer, 5!, since now the one committee member who was left over must take the new label.