When we take the Cartesian Product of two sets, does the order in which we take the Cartesian Product matter? For example if I have two sets A and B, where I take the Cartesian Product of both, does it matter if I perform the operation in this order AxB or whether I perform the operation in this order or the order BxA, or does the order not even matter? I am enquiring because I know that in both resulting sets from the Cartesian Product Operation, the ordered pairs have the same composition how every the elements of the tuples have a different order, but does this matter/ make a difference?
 A: For any two sets $A$ and $B$, there is a canonical bijection between $A \times B$ and $B \times A$, but they are not the same set, unless $A$ and $B$ are the same or one of them is empty.
A: The cartesian product is a set, and in a set as such  there is no order ( due to the extensionality principle).
So , in the cartesian product the " order" of the couples ( i.e. the elements of the cartesian product)  does not matter.
Let $A = \{a, b\}$ and $ B = \{1,2\}$.
$A\times B = \{ <a,1>, <b,2>\} = \{ <b,2>, <a,1>\}$
But the order of the elements inside the couples ( ordered pairs) themselves is essential to the cartesian product.
Take a cartesain product and inverse the order of any single couple $<a,b>$ ( with $a \neq b $) , you have totally lost your original set ( due to the extensionality principle).
A fortiori, if you inverse all the ordered pairs , the cartesian product cannot be preserved.
Note : extensionality principle : set A and set B are equal ( i.e. identical) iff they have all their elements in common.
A: The order certainly matters, namely $A=\{x\,:\, \exists y, (x,y)\in A\times B\}$ and $B=\{x\,:\, \exists y, (y,x)\in A\times B\}$. While there are instances where this could be treated as a technicality (and the longer I think about them, the fewer they become), there are others where this is a very much desired feature: namely, when you define a function $f:A\to B$ as a subset $f\subseteq A\times B$ such that for all $x\in A$ there is exactly one $y\in B$ such that $(x,y)\in f$, you demand an asymmetry between the domain $A$ and the codomain $B$, which set theorist went to considerable length to achieve.
A: They are in general different sets, so it does matter, $A\times B$ will have different elements compared to $B\times A$ . A simple example can be one where you take $A=\{0,1\}$ and $B=\{2,3\}$ then $(0,2)\in A\times B$ and $(0,2)\notin B\times A$ just like $(3,1)\in B\times A$ and $(3,1)\notin A\times B$ . Although there is a natural bijection between the two sets, just take $f:A\times B \to B\times A$ such that $f(a,b)=(b,a)$.
