Counting with combinations vs manually Lets say there is an urn containing 10 balls, 5 are red and 5 are black. 3 are drawn without replacement. I am finding the probability that I get 3 red balls.
I understand that I can count this in two different ways:
1.) (5/10)(4/9)(3/8) 
2.) 5C3/10C3
and that when I expand and simplify the second expression, it reduces to the first one. My question is that how do I intuitively recognize this? 
I realize that by expanding nCr=n!/(n-r)!r!, I am counting all the possible arrangements of n objects and then dividing by (n-r)! and r! because I don't want to count the different permutations both divisions. How can I intuitively know NOT to count this same situation with 2.) when the balls are drawn WITH replacement? 
 A: 
Find the probability that three red balls are drawn when three balls are drawn without replacement from an urn containing five red and five black balls.

You correctly found the answer in two ways.  In the first, you counted ordered selections.  In the second, you counted unordered selections.  However, this is not the best example to illustrate this, so let's consider a different example.

Find the probability that three balls of different colors are selected when three balls are drawn without replacement from a bag containing four blue, three green, and two red balls.

Ordered selections
There are $9$ ways to select the first ball, $8$ ways to select the second, and $7$ ways to select the third, so there are $9 \cdot 8 \cdot 7$ selections in our sample space.  
Let $b$ denote blue, $g$ denote green, and $r$ denote red.  There are $3! = 6$ favorable cases:  $$(b, g, r), (b, r, g), (g, b, r), (g, r, b), (r, b, g), (r, g, b)$$  Hence, the probability of selecting three different colors is 
\begin{align*}
p & = p((b, g, r)) + p((b, r, g)) + p((g, b, r)) + p((g, r, b)) + p((r, b, g)) + p((r, g, b))\\
  & = \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7} + \frac{4}{9} \cdot \frac{2}{8} \cdot \frac{3}{7} + \frac{3}{9} \cdot \frac{4}{8} \cdot \frac{2}{7} + \frac{3}{9} \cdot \frac{2}{8} \cdot \frac{4}{7} + \frac{2}{9} \cdot \frac{4}{8} \cdot \frac{3}{7} + \frac{2}{9} \cdot \frac{3}{8} \cdot \frac{4}{7}\\
& = 6 \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}\\
& = \frac{2}{7}
\end{align*} 
Unordered selections
There are $$\binom{9}{3}$$ ways to select three of the nine available balls.  
The favorable cases consist of choosing one of the four blue balls, one of the three green balls, and one of the two red balls.  Hence, the number of favorable cases is 
$$\binom{4}{1}\binom{3}{1}\binom{2}{1}$$
Therefore, the probability of selecting three balls of different colors is 
$$p = \frac{\dbinom{4}{1}\dbinom{3}{1}\dbinom{2}{1}}{\dbinom{9}{3}} = \frac{24}{84} = \frac{2}{7}$$
The reason that we obtain the same answer for ordered and unordered selections is that the order in which we select the elements of the subset does not matter, just which elements we select.

How do I know not to count the same situation with combinations if the balls are drawn with replacement?

The binomial coefficient 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
counts the number of ways of selecting a subset of $k$ elements from a set with $n$ elements.
If the balls are chosen without replacement and $k > 1$, we are not choosing a subset of $k$ balls since all the balls are available for each selection.  In particular, the same ball may be selected more than once.

Find the probability that three balls of different colors when three balls are drawn with replacement from a bag containing four blue, three green, and two red balls.  

Thre are $9$ ways to select the first ball, $9$ ways to select the second ball, and $9$ ways to select the third ball.  Hence, there are $9^3$ selections in our sample space.
The favorable cases are the same as in the preceding example.  However, the probability of selecting three different colors is 
\begin{align*}
p & = p((b, g, r)) + p((b, r, g)) + p((g, b, r)) + p((g, r, b)) + p((r, b, g)) + p((r, g, b))\\
  & = \frac{4}{9} \cdot \frac{3}{9} \cdot \frac{2}{9} + \frac{4}{9} \cdot \frac{2}{9} \cdot \frac{3}{9} + \frac{3}{9} \cdot \frac{4}{9} \cdot \frac{2}{9} + \frac{3}{9} \cdot \frac{2}{9} \cdot \frac{4}{9} + \frac{2}{9} \cdot \frac{4}{9} \cdot \frac{3}{9} + \frac{2}{9} \cdot \frac{3}{9} \cdot \frac{4}{9}\\
& = 6 \cdot \frac{4}{9} \cdot \frac{3}{9} \cdot \frac{2}{9}\\
& = \frac{16}{81}
\end{align*} 
Notice that the order of selection matters, so using combinations would not be appropriate.
