How can I see easily that redundancies are removed by dividing by the factorial? As a (stupid) example, consider the problem of choosing three pizza toppings from five toppings.
Let's suppose order matters. (Obviously, this is unrealistic. No one is going to care if it's a pepperoni, sausage, and pineapple pizza versus a sausage, pepperoni, and pineapple pizza.)
I have $5 \times 4 \times 3 = 60$ possibilities.
Now, as I've been told, dividing by $3!$ gives me the number of values with redundancies removed (i.e., order not mattering), resulting in $\binom{60}{3}$.
Is there an easy way to see/visualize this? I feel like this has just been taken for granted ever since I learned this concept.
Generally speaking, if I want to grab $r$ objects from $n$ objects, how can I easily see that
$$\binom{n}{r} = \dfrac{n(n-1)\cdots(n-r+1)}{r!}$$
results in grabbing $r$ objects from $n$ objects, with order not mattering (i.e., redundancies removed)?
 A: Sure! The magic term is equivalence classes - think about how you group together options that have different orderings, but yield the same set of toppings. 
For example, Pepperoni-Sausage-Onion (PSO) gives the same set of toppings as SPO, OPS, POS, . . . You want to bundle together the toppings patterns that are "similar" in this sense, and then count the bundles. (These bundles of similar things are called equivalence classes - two patterns are equivalent if they yield the same set of toppings.)
So how many bundles are there? Well, that's just $${\mbox{Number of toppings patterns, total}\over{\mbox{Number of patterns per bundle}}}$$ (do you see why?). So we just want to count the number of toppings patterns that are "similar" to each other. That is:

How many ways are there to get the same set of toppings as Pepperoni-Sausage-Onion?

Do you see why the answer to this is $3!$?

Now here's a more difficult problem - let's say I'm allowed to repeat ingredients! So I could order a PSO pizza, or a PSS pizza.
Now things get messy. First, the number of total toppings patterns changes: it's $5\cdot5\cdot5$ now (do you see why?). That's a minor issue, though.
More importantly, the equivalence classes have different sizes! There are six patters equivalent to PSO (note that "PSO" itself is one of those six), but only three patterns equivalent to PSS. So the picture isn't going to be as simple as "total number of patterns, divided by redundancy" - we need to do some more work.
I'm going to leave this problem as an exercise - it's fun to work out on your own. Here's a hint though: how many patterns are there with six equivalent versions? How about three? How about one? 
