Cartesian product (without order) and operation

I think that I know the formal definition of the Cartesian product of order pairs, but in my example the order doesn’t make sense. Let’s look at an example. I have 4 sets: $A_1=\{ milk, 3\}$, $A_2=\{water, 1\}$, $A_3=\{black, no, 1 \}$, $A_4=\{milk, no \}$

The Cartesian product $A = A_1 \times A_2 \times A_3 \times A_4 = \{$

$( milk , water , black , milk )$,

$( milk , water , black , no )$,

$( milk , water , no , milk )$,

$( milk , water , no , no )$,

$( milk , water , 1 , milk )$,

$( milk , water , 1 , no )$,

$( milk , 1 , black , milk )$,

$( milk , 1 , black , no )$,

$( milk , 1 , no , milk )$,

$( milk , 1 , no , no )$,

$( milk , 1 , 1 , milk )$,

$( milk , 1 , 1 , no )$,

$( 3 , water , black , milk )$,

$( 3 , water , black , no )$,

$( 3 , water , no , milk )$,

$( 3 , water , no , no )$,

$( 3 , water , 1 , milk )$,

$( 3 , water , 1 , no )$,

$( 3 , 1 , black , milk )$,

$( 3 , 1 , black , no )$,

$( 3 , 1 , no , milk )$,

$( 3 , 1 , no , no )$,

$( 3 , 1 , 1 , milk )$,

$( 3 , 1 , 1 , no ) \}$

Is the above right? This is my first question but next refers to operations. Let’s take a subset $$B_1=\{(milk,1,1,milk) \}, B_2=\{ (milk,1 ,1,no)\}, B_3=\{(milk,1,1,milk), (milk,1,1,no) \}$$ I think that every tuple out of 24 in set A is an element that doesn't change at all. I mean that $C_1=\{ (milk,1 ,1,no)\}$ and $C_2=\{ milk,1 ,1,no\}$ differs because $C_2=\{ milk,1,no\}$. If so:
$$B_1 \cap B_2 = \emptyset; B_1 \cup B_2 = B_3; B_1 \cap B_3 = B_1; B_1 - B_2=B_1; B_3 - B_2=B_2; B_1 - B_3= \emptyset; B_1 \cup B_2 \cup B_3 = B_3$$

Does anyone could confirm (or not) operations truthfulness?

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Yes, what you have written appears to be the the Cartesian product $A_1 \times A_2 \times A_2 \times A_4$.
Your statements about your subsets $B_1$, $B_2$, and $B_3$ of the Cartesian product are all correct except that $B_3 - B_2$ is $B_1$, not $B_2$.
Trevor: thank you. You are absolutely right about order. Order within every 4-tuples (in the round braces) is crucial, but re-arranging 4-tuples within the curly braces can be done without limits. The last: of course $B_1$ - my stupid mistake. – user65517 Mar 7 '13 at 20:50