An instructor has an exam of 5 questions and a question bank of 50 questions. Are there $50\choose 5$ or $(50)_5$ possible exams? Where $(50)_5 = 50! / 45! $ and $50\choose 5$ $= 50!/(5!45!)$ 
I've narrowed it down to these two possibilities (though even these may be wrong), but I think I'm confusing myself as to which one I use. Can anyone suggest a good way of thinking of this problem that will make it apparent whether I need use use the falling factorial or the binary coefficient? 
Edit: thank you for all the answers! Indeed, I meant $50!/45$ as the falling factorial. Also, the question doesn't specify whether it was meant to be ordered or unordered, but I think I understand my confusion now!
 A: This depends.
If the order of the question matters (i.e. if two exams with the 5 same questions but in another order are considered different), there are $50.49.48.47.46$ possibilities. 
If the order of the questions does not matter, it is $\binom{50}{5}$.
A: To reiterate and emphasize, this entirely matters on what you consider as being considered "different" versus "the same."  And, in the long run, it is possible that such a distinction doesn't even matter in the first place!
Suppose we were asking this question in order to solve the probability question of "A student had time to study $30$ out of $50$ possible exam questions.  If the exam contains five randomly selected exam questions from those $50$, what is the probability that the student had studied each of those questions?"
I must emphasize, we have a choice of what sample space to use to describe this problem.  There are many different sample spaces that we could have chosen to describe this, and whichever we choose is entirely up to us!  There are certain benefits to choosing certain sample spaces though compared to others.  In particular, some things to look out for are making sure the events we are interested in calculating the probability of are valid events in our sample space, simplicity of the sample space, and whether or not the sample space is such that each outcome in it is equally likely to occur.  After all, we may only use counting techniques to find probability by setting $Pr(A)=\frac{|A|}{|S|}$ in equiprobable sample spaces.

In solving the above question with the sample space being that we treat the order that the questions appear as relevant, we would have $30\cdot 29\cdot 28\cdot 27\cdot 26$ possible exams where our student studied every question out of $50\cdot 49\cdot 48\cdot 47\cdot 46$ possible exams.
In solving the above question with the sample space being that we treated the order that the questions appear in as not relevant, we would have $\binom{30}{5}$ possible exams where our student studied every question out of $\binom{50}{5}$ possible exams.
Taking the ratio in each case gives us the probability.  Notice that
$$\dfrac{30\cdot 29\cdot 28\cdot 27\cdot 26}{50\cdot 49\cdot 48\cdot 47\cdot 46} = \dfrac{\binom{30}{5}}{\binom{50}{5}}$$
and so it did not matter which interpretation we used for this as both approaches led to the same answer.  Of course, if we were to then ask "What is the probability that the first question on the test was one that the student studied" then the sample space where order does not matter cannot be the one that we use as it did not keep information about which spot in the exam was occupied by which question.  But, then again, we aren't restricted to using the first sample space either and we could have considered the sample space which only kept track of which question was the first question, giving an answer of $\dfrac{30}{50}$, which is simpler than using the first which would have given an equal but more tiresome answer of $\dfrac{30\cdot 49\cdot 48\cdot 47\cdot 46}{50\cdot 49\cdot 48\cdot 47\cdot 46}$.
A: If the order of questions on the test does not matter, the answer is $$\binom {50}{5}= \frac {50!}{5!45!}=2118760$$ 
Otherwise we have $$ \frac {50!}{45!}=254251200$$ many tests. 
