How do I know when to take ordering into account? 
I'm quite confused as to why in (c) the ordering matters but not in (b)
I kind of understand in (a) ordering doesn't really matter since all the numbers are the same therefore switching places doesn't really make any difference. 
However, for (c) the solution is showing that $3* \frac{1}{6} * \frac{5}{6}$ I assume it means that we have to take ordering into account. That means  $5,1,5$ is different from $5,5,1$.
Following the same logic, for (b) why don't we also take ordering into account? Because clearly $6,2,3$ is different from $2,3,6$, right?
What am I missing here?
 A: I think your confusion comes from the following.  The answer to (b) is 
$$\frac{\binom{6}{1}\binom{5}{1}\binom{4}{1}}{6^3} = \frac{5}{9},$$
and probably you are wondering why it is not
$$3\frac{\binom{6}{1}\binom{5}{1}\binom{4}{1}}{6^2},$$
which "takes ordering into account."
Apart from the fact that the latter probability is greater than $1$, you can see that the "ordering" is already taken into account in the first answer if you consider a simpler example.  Suppose that there are two hypothetical dices, each having only three sides $\{1,2\}$ and each side showing with equal probability.  Four outcomes are possible:
$$(1,1), (1,2), (2,1), (2,2).$$
The probability of showing different numbers is clearly
$$\frac{\binom{2}{1}\binom{1}{1}}{2^2} = \frac{1}{2}.$$

Note: you can easily obtain an answer to (b) once you know answers to (a) and (c).

A: It helps to distinguish between the dice.  Suppose the dice are, respectively, blue, green, and red.  Then an outcome can be represented by the ordered triple $(b, g, r)$, where $b, g, r \in \{1, 2, 3, 4, 5, 6\}$.

I roll three six-sided fair dice.  What is the probability that all three dice show the same number?

Since there are six possible outcomes for each die, there are $6^3$ possible outcomes in the sample space.
Of these, just six show the same number on each die. 
Hence, 
$$\Pr(\text{same number on each die}) = \frac{6}{6^3} = \frac{1}{6^2} = \frac{1}{36}$$

I roll three six-sided fair dice.  What is the probability that each die shows a different number?

The sample space is the same as above.
For each of the six possible numbers on the blue die, there are five possible numbers on the green die that are different from the one on the blue die, and four possible numbers on the red die that differ from both the number on the blue die and the number on the green die.  Hence, there are $6 \cdot 5 \cdot 4$ favorable outcomes.
Hence,
$$\Pr(\text{different number on each die}) = \frac{6 \cdot 5 \cdot 4}{6^3} = \frac{5 \cdot 4}{6^2} = \frac{5}{9}$$

I roll three six-sided fair dice.  What is the probability that all two dice show one number and the other die shows a different number?

Again, the sample space is the same as above.
There are $\binom{3}{2}$ ways to select which two of the three dice show the same outcome.  There are six possible numbers for both of those dice to show.  For each of these possibilities, there are five possible numbers the third die could show.  Hence, the number of favorable outcomes is 
$$\binom{3}{2} \cdot 6 \cdot 5 = 3 \cdot 6 \cdot 5$$
Hence, 
$$\Pr(\text{two dice show the same number and the other die shows a different number})\\
 =  \frac{3 \cdot 6 \cdot 5}{6^3} = \frac{3 \cdot 5}{6^2} = \frac{5}{12}$$
Since these three cases are mutually exclusive and exhaustive, their probabilities must have sum $1$, which you can verify by direct calculation.
Note:  The reason for distinguishing between the dice is to ensure that all outcomes are equally likely.  For instance, if all three dice were white, the outcome $(1, 2, 3)$ would occur $3! = 6$ times as often as $(1, 1, 1)$.  
A: Order is considered in each numerator.   We count the ways to select dice for singles, pairs, or tripple as appropriate, and the distinct numbers to show on them.
Note that $6^3 = \tbinom 3 {3}\binom 6{1,5}+\tbinom 3 {2,1}\binom 6{1,1,4}+\tbinom 3{1,1,1}\tbinom 6{3,3}$ where each term counts ways to select a tripple, a pair-and-single, and three distinct singles, respectively.
That is, $\tbinom 33$ counts the ways to select 3 from 3 dice to be identical (which is of course 1), and $\tbinom 6{1,5}$ or $\tfrac {6!}{1!5!}$ counts the ways to select one from six faces for those dice (which is of course 6).   Thus the probabiliy for obtaining all identical faces is $1/36$
Likewise, $\tbinom 3{2,1}$ counts ways to select which 2 from 3 dice might be a pair (which is 3), and $\tbinom 6{1,1,4}$ counts ways to select two from 6 faces to be the pair and single (which is $6\cdot 5$ or $30$).   Thus the probability for obtaining a pair and single is $15/36$.
And $\tbinom 3{1,1,1}$ counts ways to select dice to be three singles (which is 6), and $\tbinom 6{3,3}$ counts ways to select the three distinct faces to show on those dice (which is $20$).   Thus the probability for obtaining a pair and single is $20/36$.

