A student must pick 5 classes from 12 courses, if he must have at least one WH class or USH class, how many different choices does he have? The World History course and United States history course is already part of the 12 courses he must choose from. It is required to have at least one of these classes in his new class schedule. Sorry if my question didn't make sense earlier, this question was from a practice exam I'm just recalling the question from memory. I was short on time so I just solved C(12,5) and I know its wrong because I forgot about the at least. This has been bugging me all day and I'd appreciate it if someone can explain. 

Edit: this answer is solved for a different problem. Specifically I was under the presumption all classes were mandatory rather than only one class being mandatory. I will leave it be however. 
Short answer: hold 2 classes fixed for C(10,3)
Tall answer:
The answer C(12,5)-C(10,5) is wrong. Counterexample: 2 classes to choose among 4 with 1 biology and 1 world history has one solution. However C(4,2)-C(2,2)=5.
This answer does not remove the combinations C(2,1)C(2,1) which correspond to only one class being chosen.
Also wrong is the Star's and Bar's approach as it presumes there are infinite biology and world history classes. The wording "at least" is very suggestive, however the problem cannot be solved without knowing the actual number of duplicate classes as is seen from observing a simpler case (a technique you should pick up).
The general formula for choosing k classes among n, with m classes being mandatory and not having duplicates is C(n-k,m-k) because you are holding m classes fixed. However following the other line of reasoning of counting up all combinations, subtracting those lacking 1 mandatory class, those lacking 2 mandatory classes,  etc... 
$$C(n,k)-C(m,m-1)C(n-m,k-m+1)-C(m,m-2)C(n-m,k-m+2)-\cdots-C(m,0)C(n-m,k) = C(n-m,k-m)$$
Because
$$\sum_{i = 0}^m C(m,i)C(n-m,k-i)= C(n,k)$$ and is a well known result in combinatorics
