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.
 A: When you define a set by giving the list of its elements, duplicates do not matter, i.e. the number of times you repeat an element is irrelevant, as well as their order.
The only important thing is to determine whether  an object occurs (at least one time) or not in your list.
Therefore,
\begin{align}
 \{1,1,2\} &= \{1,2\}
\\
 \{3,3,3,2,2,4\} &= \{2,3,4\}
\end{align}
This "principle" is true also for the cartesian product, because it is a set. So,
\begin{align}
\{(1,3),(1,2),(1,4),(2,3),(2,2),(2,4),(1,3), (1,3),(1,4))\} &= \{(1,3),(1,2),(1,4),(2,3),(2,2),(2,4)\}
\end{align}

From a theoretical point of view, this approach is justified by the so called axiom of extensionality, which states that two sets are equal if and only if they contain the same elements. 
A: No, because the Cartesian product of sets is itself a set. For sets in general, we consider a set, and a set with the same entries but some duplicates, to be precisely the same.
For example, let $A=\{1,2\},B=\{3,4\},A'=\{1,1,2,2\},B'=\{3,3,4,4,4,4,4\}$. Under these conditions, $A=A',B=B',$ and in turn $A \times B = A' \times B'$.
Thus, the inclusion of duplicate elements does not matter here. Why you would want to include them is beyond me, unless you're working with objects like multisets (which do admit duplicate elements). In such an instance, obviously the opposite holds true. But unless stated otherwise I imagine the first instance holds for you, i.e. you should be removing duplicates from sets.
A: 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). 
