counting another problem I am trying to do my homework and it seems really hard. i would like to get checked here and make sure that im on the right track. can anyone help me??
Question: A group of hundred students want to create a committee of twelve which will then select a chairman for the committee.
a) In how many ways can this be accomplished?
b) what if they decide to have two members serve as co-chairs??
Answer a) $P(100,12)= \dfrac{100!}{12!(100-12)!}$;
Answer b) $P(100,12)+ P(12,1)+ P(12,2)$.
since the order matters and repetition is not allowed, i selected 12 from 100 students.
did i do it right??
 A: Your answer to a is correct for selecting the committee, but ignores choosing the chairman.  The use of $P(100,12)$ is strange, because the order of the selection does not matter.  Your expression would usually be written $C(100,12)$ or ${100 \choose 12}$ with the value you give $\frac {100!}{12!(100-12)!}$.  I would take $P(100,12)$ to say that order matters and evaluate it as $\frac {100!}{(100-12)!}$.  Now you need to multiply by $C(12,1)={12 \choose 1}$ to pick the chairman.  
For b, you should multiply by the number of ways to select the co-chairs.  It seems order does not matter here (it is not a chair and vice-chair, where order would matter).
A: 
A group of hundred students want to create a committee of twelve which will then select a chairman for the committee. In how many ways can this be accomplished?

First you're choosing 12 out of 100 people to be on the committee: $\binom{100}{12}$. Then you're choosing one out of that 12 to be the committee head: $\binom{12}{1}$.

what if they decide to have two members serve as co-chairs??

Same as before, only now instead of choosing a single person for committee head, you're choosing two: $\binom{12}{2}$.
