Discrete Maths: Ways of sitting 3 exams during a 13 week semester. I had a discrete maths test today and this question was the last one and it really threw me off.

During a 13 week semester ( no breaks in between ), a student must sit 3 exams >for a particular course. 
  However, he must sit them in order, i.e 1st -> 2nd -> 3rd. In other words, you must first sit the 1st exam, then the 2nd then the 3rd. so can't sit the second exam without having sat the 1st one, etc..
If he can only sit one exam per week, in how many ways can the student sit the exams.

I tried during inclusion/exclusion principle but it got messy real quick. I also thought of doing the lines and dots method but figured it's wrong since the dots are not identical.
I'm just curious how to do this.
the second part of the question was

Suppose they are not allowed to sit two exams in consecutive weeks, now in how >many ways can the student sit the exams.

I didn't even attempt the second part as I did not have time.
Any help would be greatly appreciated. 

(Edited the answer to fit all criterias.)
Choose the 3 weeks out of 13 on which the student writes the exam. On the first chosen week the first exam, the second chosen week the second exam, third week the third.
There is a known mathematical formula for choosing $k$ instances from $n$, which is calculated by
$${n\choose k} = \frac{n!}{k!(n-k)!}$$
Note that this already fits the criteria of writing the exams in order. It just selects $3$ weeks out of $13$, and on the $n^{th}$ selected week ($n \in \{1,2,3\}$) you write the $n^{th}$ exam.
In this case:
$${13\choose 3} = \frac{13!}{3!(13-3)!} = 268$$
However, if you need to have at least $1$ week off before you write the next exam, you really only have $11$ objects to order (instead of $13$):


*

*(First exam's week + a week off)

*(Second exam's week + a week off)

*(Third exam's week)

*(Other exam free weeks)$\times 8$


You need to choose $3$ places out of $11$ to fit these weeks (or double weeks) into the semester, so the answer is:
$${11\choose 3} = \frac{11!}{3!(11-3)!} = 165.$$
