Dice outcomes probability 
Two dice are rolled. What is the probability that the sum of the
  numbers on the dice is at least 10

Let $Z$ denote the set of successful outcomes:
$Z=\{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\}\\
\text{Sample space: }
S=6^2$
So the answer gives the probability as, $P=\frac{6}{36}$.
My question is why are the pairs $(5,5),(6,6)$ only included once? The way I see it for two dice $D_1,D_2$ is it may appear as $D_1: 5,D_2: 5$ or $D_2: 5,D_1: 5$, so why don't we count for this as two outcomes?
 A: Let $(x,y)$ be the ordered pair of rolls.  When $x\ne y$, we have $(x,y)\ne (y,x)$.  For instance, $(3,5)$ and $(5,3)$ are different.  However, when $x=y$, we have $(x,x)=(x,x)$, (even though I switched the $x$s).  
You may want to consider the table of two die rolls that shows all possible outcomes:
$$\begin{array}{c|c|c|c|c|c|c|} &1&2&3&4&5&6 \\
 \hline 1&(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6) \\ 
 \hline 2&(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6) \\
&&&etc.\end{array}$$ 
Each double $(x,x)$ occurs only once in the table.
A: Your sample space $S=\{1,2,3,4,5,6\}^2$ has exactly 36 unique outcomes, with each having probability $\frac{1}{36}$. $(4,6)$ and $(6,4)$ are separate outcomes, whereas $(5,5)$ and $(5,5)$ are the same.

Intuitively, to get an outcome involving one $4$ and one $6$, I could roll either a $4$ or a $6$ for the first die, which has a probability of $\frac{2}{6}$ of success. If I was successful with the first die, I have a $\frac{1}{6}$ chance of rolling the number I'm missing with the second die. The total probability of success here is $\frac{2}{6}\cdot\frac{1}{6}=\frac{2}{36}$.
On the other hand, if I want an outcome with two $5$'s, I need to roll a $5$ on the first outcome, with probability $\frac{1}{6}$ of success. Given I succeed, I need to roll a $5$ again with the second die, with probability $\frac{1}{6}$. The total probability of success here is $\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$.
A: The union of two sets $A$ and $B$ is defined as 
$$ A\cup B = A + B - A\cap B $$
which prevents adding any items which are in common to both sets twice.
For your specific question, $(5, 5)$ and $(6, 6)$ are common to both sets and thus are not counted again. If you counted them both ways, as you suggested in your answer, you would find yourself stuck with the fallacy of "double counting".
For a more rigorous explanation of this concept in combinatorics, feel free to look up the "Inclusion-exclusion principle".  
