Figuring out number within event and sample space 
Exercise 1.8: Suppose that a bag of scrabble tiles contains 5 Es, 4 As, 3 Ns and 2 Bs. It is my turn and I draw 4 tiles from the bag without replacement. Assume that my draw is uniformly random. Let $C$ be the event that I got two Es, one A, and one N. 
a) Compute P($C$) be imagining that the tiles are drawn one by one as an ordered sample. 
b) Compute P($C$) by imagining that the tiles are drawn all at once as an unordered sample.

How do go about differentiating between the different situations and apply the combinatorics tools to count the cardinality of each event and sample space?
So far I know:
$\binom52$ := Number of ways to choose Es; $\binom31$ := Number of ways to choose A; $\binom21$:= Number of ways to choose N.
Moreover, there are 10 total permutations of the set $\{E, E, A, N\}$, where we do not distinguish between the Es. 
Given that where does the order vs unordered drawing of tiles impact the exercise? Is it in the determination of the sample space or in the event cardinality. I'm assuming that it'll effect mainly the cardinality of the sample space, with it being $\frac{14!}{10!}$ for a, and $14^5$ for b. 
Any hints, or insight would be helpful!
 A: In order to adequately explain... let us temporarily imagine that our scrabble tiles are all uniquely numbered.  This will allow us to distinguish between otherwise indistinguishable outcomes and now work in a sample space where every outcome is equally likely to occur as every other outcome.
Now... suppose our tiles were $\{E_1,E_2,E_3,E_4,E_5,A_1,A_2,A_3,A_4,N_1,N_2,N_3,B_1,B_2\}$
As we are drawing tiles without replacement, for part (a) we have the following example outcomes: $(E_1,N_1,N_2,B_1), (E_1,N_2,N_1,B_1), (B_1,N_1,N_2,E_1),\dots$ and these all count as being different since the order in which they occurred is different.
By applying the rule of product, we can count how many different outcomes possible there are in the sample space as being $14\times 13\times 12\times 11$.  This, since there are $14$ options for what the first tile was that we drew, $13$ options for what the second tile was, and so on... multiplying the results to finish the count.  This is equal to what you proposed for the count of the sample space in your question.
As for trying to count how many of these outcomes correspond to having two $E$'s, one $A$ and one $N$... we can do so by first picking which of the four positions was occupied by an $A$, then which specific $A$ it was for that position, which of the remaining positions was occupied by an $N$, then which specific $N$ it was in that position, then for the left-most remaining space choose which specific $E$ and finally pick which $E$ for the final position.  This gives a count of $4\times 4\times 3\times 3\times 5\times 4$.
Finally, recalling that for an equiprobable sample space we can calculate probabilities by taking the count of good outcomes divided by the total number of outcomes, we arrive at a probability of:
$$\frac{4\times 4\times 3\times 3\times 5\times 4}{14\times 13\times 12\times 11}=\frac{120}{1001}\approx 0.12$$

For part (b), we are drawing the tiles simultaneously without replacement.  We find then that there are $\binom{14}{4}$ different outcomes.  Your attempt of $14^5$ was off... $14^5$ is the number of outcomes if order mattered and there was replacement and we were drawing five tiles instead of four.  Here, we are still only pulling four tiles, not five, order doesn't matter, and there is not replacement.
Among the possibilities are things like $\{E_1,N_1,N_2,B_1\}$, the fact that it is a subset of the set of tiles being emphasized by the curly brackets rather than parentheses, noting that rearranging the terms within a subset does not make the subset "different."
Now, the number of outcomes in this sample space corresponding to our desired event, we choose which $E$'s appeared (simultaneously since order doesn't matter), which $A$ appeared, and which $N$ appeared.  This gives a count of $\binom{5}{2}\times 4\times 3$.  Remember, since order doesn't matter to us here, we don't bother caring about permuting the positions of the tiles.
Again, recognizing that each of the outcomes in the sample space are equally likely, by dividing the number of good outcomes by the total number of outcomes this gives us the probability of our event as being:
$$\frac{\binom{5}{2}\times 4\times 3}{\binom{14}{4}}=\frac{120}{1001}\approx 0.12$$

What is the takeaway from this?  We can in many problems choose for ourselves whether or not order matters for problems of calculating probability as it does not in any way affect the answer so long as the event we are interested in calculating the probability of does not make any reference to specific order of outcomes.  It is a choice as to what sample space to use.  The decision as to which sample space to use should be dictated primarily by making sure that it adequately describes the event(s) you are interested in calculating probabilities of, then trying to make sure that your choice of a sample space is one which is equiprobable if you plan on using counting techniques to continue.  After that, picking whichever you are most comfortable with or whichever makes the arithmetic easiest in the end for you.
In my experience, for problems like the one you post or poker problems etc... it is easiest in my opinion to treat it where order didn't matter, but it is not incorrect to treat it as though order did matter... so long as you account for order correctly in both the numerator and denominator.
