Confusion regarding the scope of multiplying a factorial by a constant $30(3k)!$ I've stumbled upon this factorial, which is written exactly in this way:
$$30(3k)!$$
As simple as this may seem, I honestly don't know how to interpret this. Does the factorial encompass the entire thing to give me $(30\cdot 3k)! = (90k)!$ ? Or would the $30$ multiply the final product of $3k$ like so: $30(3k!) = 90 \cdot k!$?
My last question is does this difference really matter when it comes to factorials?
 A: This question basically comes down to understanding where the factorial functions fits into the "order of operations."  The first thing to be aware of is that the order of operations is a convention, and not a mathematical law or a consequence of the axioms.  It is something which we have agreed on so that we can use notation consistently.
With that in mind, the usual convention is that the factorial takes precedence over multiplication.  Thus the correct interpretation is that
$$ 30(3k)!
= 30 [(3k)!]
= 30 (3k)(3k-1)(3k-2)\dotsb(3)(2)(1). $$
As David G. Stork points out in his comment, this is probably the only reasonable interpretation of the notation as, otherwise, the expression $5!3!$ is likely to be quite ambiguous.
Of course, as indicated by the notation above, additional grouping symbols can also make life easier.  You could also change the order in which you write your symbols.  For example,
$$ (3k)! \cdot 30 $$
is less ambiguous that the notation you give.
Some additional commentary:  the factorial function is one of the few examples in standard mathematical notation of a postfix operator.  Usually, when we have a function which maps some set to another, we prepend that function to the object upon which it acts.  For example, if a function $f$ acts on some object $x$, then we may denote this by $f(x)$.  With this notation, we can write things like
$$ f(x) 10
\qquad\text{or}\qquad
4 f(x)^5 $$
without ambiguity.  The factorial notation should be interpreted the same way, though we have placed the symbol for the function on the right rather than the left.  That is, we could reasonably write
$$ ! : \mathbb{N} \to \mathbb{N}
\qquad\text{is defined by}\qquad
 n \mapsto (n)! = n(n-1)(n-2)\dotsb(3)(2)(1). $$

As to your last question, it absolutely matters.
$$ (3k)! \cdot 30
\qquad\text{is quite a lot smaller than}\qquad
(30 \cdot 3k)!. $$
For example, if $k=1$, note that
$$ (3\cdot 1)!\cdot 30 = (3)(2)(1)(30) = 180
\qquad\text{while}\qquad
(30 \cdot 3 \cdot 1)! = 90! = \text{something very big}. $$
