Would duplicates matter in cartesian product of a set? For example:
\begin{align}
A &= \{1, 1, 2\}
\\
B &= \{3, 3, 3, 2, 2, 4\}
\end{align}
Would $A$ cross $B$ equate to $\{(1,3),(1,2),(1,4),(2,3),(2,2),(2,4)\}$ without the dupes of $(1,3)$, etc.

Consider the sets 
$\{(1,3),(1,2),(1,4),(2,3),(2,2),(2,4)\}$
and 
$\{(1,3), (1,3), (1,3), (1,3), (1,3), (1,3), (1,2),(1,4),(2,3),(2,2),(2,4)\}$
Can you see an element of one set that is not also an element of the other? ( The elements being here ordered pairs). 
In case you answered no, you are entitled to say that the "two" sets are, in fact, one and the same set. 
The reason is that the extensionality principle 
$\forall (x) ( x\in A\iff x\in B)$
is equivalent to 
$\neg\exists (x) [ (x\in A \land x\notin B) \lor ( x\in B \land  \notin A) ] $
( In words : there is no $x$ such that , $x$ is in $A$ but not in $B$, or, $x$ is in $B$ but not in $A$). 
which is equivalent to 
$A \Delta B = \emptyset$. 
( The symmetric difference of set $A$ and set $B$ is empty). 
