All you care about is the order of $A,B,C$. There are $3!=6$ orders of them,of which $2$ have $C$ in the middle. So $\frac 26=\frac 13$ of the arrangements have $C$ in the middle. $\frac {6!}3=240$
I'm not sure there's a reductionistic trick that you can apply that will work generally. One learns the primitives well enough, and by well enough, I mean that one knows not only what each of them does, but what they can do in combination. With enough facility at that, one can construct any number of combinatorial frameworks to solve general problems.
In the kind of problems that you refer to specifically, one way to approach the problem is to consider how one would construct the selection. How would one go about constructing an eight-card hand that contains at least one heart? Well, you could take one of the $13$ hearts, and seven of the remaining $39$ cards; or two of the $13$ hearts, and six of the remaining $39$ cards; or three of the $13$ hearts, and five of the remaining $39$ cards; and so forth. That would lead to the summation
$$
\binom{13}{1}\binom{39}{7} + \binom{13}{2}\binom{39}{6} + \cdots
+ \binom{13}{8}\binom{39}{0}
$$
Working through those terms would indeed give you the correct answer, $691014402$. However, upon looking at this, one might see a faster way to the answer, by constructing the hand "in the negative": that is to say, by finding which eight-card hands do not qualify, and subtracting them from all eight-card hands. That leads one to the simpler calculation
$$
\binom{52}{8} - \binom{39}{8}
$$
which yields the same answer of $691014402$. One could view that as a trick, but it's really just a tool. One sees which "end" of the problem is more "finite", and generally prefers to tackle the problem from that end.
Hope this helps.
Best Answer
In your first example you are choosing slots for each student to occupy, but you don't care which order the students make in each slot. You only care about the combination.
In permutations the order of the objects matters. If the objects are indistinguishable then it is a combinations problem.
In your second example you can distinguish one student from another; therefore it is permutations problem, so switching the order of the students in the line will affect the way the photograph looks.
To tell the difference you look at the 'object' in question and ask yourself is it distinguishable or not in the context of the question.