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.
The thing with dice games is that we often want to know the odds of getting a particular event ('I have a pair of sixes').
Your result is correct for calculating the number of possible combinations, with interchangeable dices. However, if we want to find a result about, say, the sum of the dices, this way of counting cases is not particularly useful since there are less chances to get the combination (6,6,6,6,6) than the combination (1,2,3,4,5).
Your book calculates the number of 'outcomes' as if each dice is of a different colour (hence, not interchangeable) and getting (1,2,3,4,5) is not the same as getting (2,3,4,5,1). The main point of this approach is to have outcomes that all have the same odds of occuring (namely, $1/6^5$). Then you can solve a lot of probability questions (odds of at least one 6, odds of sum>20, etc) just by counting the favourable cases.
Best Answer
In a permutation you have only one each of the available elements. So if you are taking permutations of three elements out of $\{1,2,3,4\}$ some of the legal ones are $(1,2,3), (3,1,4), \text {and} (2,3,1)$. There are a total of $24$ of them. As you say, combinations do not care about order. There are only four taking combinations of three out of the same set: $\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}$. Combinations with repetition would allow multiples, so $\{1,2,2\}$ becomes legal, but is still the same as $\{2,1,2\}$. These are called multisets.