It's easier to see how this works with smaller numbers. Consider 
. We can cancel like this:
This last expression is much easier to deal with. We can carry out more cancellations as we find them, and end up with an integer at the end. A more compact way of writing this out is this:
In your case, we can write:
You can keep canceling from here, as each number on the bottom has a multiple on top, or you could probably use your calculator at this step. The number 
is not too big for most calculators to handle, at 
digits. The numerator is 
digits, which is getting bit unwieldy... Cancellation is better.
Videos
I'm in discrete right now and my problem is 420! / 385!10! . Using the choose function ( n! / (n-r)! r! ) i understand how to divide when the bottom adds to the top (ex. if it was 420! / 410! 10! i would just do 420C10 (C = choose) ) Also i figured out how to do it if it adds the other way (ex. 270! / 255! 20! would become (270C10) * (1/251*252*...*255) ) However I am having trouble learning the other way since in this problem 385 and 10 dont add up to or above 420. I tried doing (420! / 410! 10!) * (1/385 *386) ... however this doesnt make sense in this problem since it is below and not multiplying on extra numbers. If someone could help me understand how to figure this out it would be great!
It's easier to see how this works with smaller numbers. Consider . We can cancel like this:
This last expression is much easier to deal with. We can carry out more cancellations as we find them, and end up with an integer at the end. A more compact way of writing this out is this:
In your case, we can write:
You can keep canceling from here, as each number on the bottom has a multiple on top, or you could probably use your calculator at this step. The number is not too big for most calculators to handle, at digits. The numerator is digits, which is getting bit unwieldy... Cancellation is better.
This is the binomial coefficient
It equals
and one can cancel everything in the denominator from the numerator.
It will leave a reasonably practical multiplication.
I'd prefer not to have to actually do it. 
HINT:
Example, set $n=5$:
$$\frac{(5+1)!}{(5+2)!}=\frac{6!}{7!}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{7}\cdot\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{7}\cdot1=\frac{1}{7}=\frac{1}{5+2}$$
For an algebraic proof:
$$ \require{cancel} \frac{(n+1)!}{(n+2)!}=\frac{(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}{(n+2)(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{\cancel{(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}}{(n+2)\cancel{(n+1)n(n-1)(n-2)(n-3)(n-4) \ldots 5\cdot 4\cdot 3\cdot 2\cdot 1}}=\frac{1}{(n+2)}$$
For example, is 5*3! = 30 or 15!? Is it considered multiplication, and done from left to right (meaning that the above example is 15!)?