You might consider whether the 5 of hearts is different from the heart 5 ? There is clearly a bijection between Rank x Suit and Suit x Rank. So A x B is in that sense quite similar to B x A.
What differentiates A x B from B x A is better revealed when A and B have the same elements (rather than for example suits and ranks), but meaning is attached to order of the pairing of the elements of A and B. What about $R^2 = R \times R$, identifiable with the x-y plane ? Now, we can see that $(x, y) \ne (y, x)$ unless $x = y$.
In the most common (ZF) treatment of ordered pairs, for $a \in A $ and $b \in B$ the ordered pair $(a, b) $ is defined as the set {a, {a, b}}. This distinguishes it from the set {b, {a, b}} unless a = b.
In More Detail
In the card example we create sets representing suits and ranks which are disjoint (in English), so that Suits = {C, D, H, S} and Ranks = {2, 3, 4, ,5, 6, 7, 8, 9, 10, J, Q, K, A}. There is a set representing cards, perhaps identified as Cards = {2C, 3C, ...AS} (at my bridge club they have bar codes for the dealing machine). We see there is a bijection between Suits x Ranks and Cards so that a card is identifiable by its suit and rank, and equally there is a bijection between Ranks x Suits and Cards and a card is identifiable by its rank and suit. Mathematically (in ZF set theory), the elements of the Cartesian products Suits x Ranks differ from the elements of Ranks x Suits: they both consist of elements which are sets, but the one looks like {C, {C, 2}} while the other looks like {2, {2, C}}, or in the more conventional ordered pair notation (C, 2) and (2, C). The elements which comprise the two Cartesian products are different so that in set terms, Suits x Ranks $\ne$ Ranks x Suits. They are interpreted to mean the same because:
- When we see an ordered pair the fact that Suits and Ranks are disjoint lets us recognize which element of the pair is a suit and which is a rank.
- There are bijections from both of the Cartesian products to Cards and we interpret the ordered pair as a card.
But, in maths a lot of ordered pairs we meet are numbers. In the X-Y plane X = {x|x $\in$ R} and Y = {y|y $\in$ R}, so in fact X = Y = R and it follows that X x Y = Y x X = R$^2$ (any pair of numbers e.g. (5, 3.72) exists in R$^2$, is a point in the X-Y plane and a point in the Y-X plane: the three Cartesian products contain the same elements and are therefore the same set). Now to understand the ordered pair we cannot identify which element belongs to which set by its value, we need to know what the order represents, i.e. the first element is a X-value, the second a Y-value.
This leads to a somewhat ironic conclusion that
- if A and B are different sets then A x B $\ne$ B x A (in set terms), but we can probably interpret the meaning of (a, b) and (b, a) to be the same thing
- if A and B are the same set then in fact A x B = B x A (in set terms), but (a, b) doesn't mean the same as (b, a) unless a = b.
Say $S_1 = \{a,b\}$ and $S_2=\{x,y,z\}$. Then
$$
S_1\times S_2= \big\{ (a,x),\ (a,y),\ (a,z),\ (b,x),\ (b,y),\ (b,z) \big\}.
$$
So you can look at it in either of two ways:
$$
\big\{ \underbrace{(a,x),\ (a,y),\ (a,z)},\ \underbrace{(b,x),\ (b,y),\ (b,z)} \big\}.
$$
$$
\big\{ \underbrace{(a,x),\ (b,x)},\ \underbrace{(a,y),\ (b,y)},\ \underbrace{(a,z),\ (b,z)} \big\}
$$
The first is a sum of two $3$s; the second is a sum of three $2$s. So the first is $2\times3$ and the second is $3\times 2$. That's one way of knowing that $2\times3$ is the same number as $3\times2$.
$$
|S_1|\times|S_2| = \sum_{a\in S_1}\ \sum_{x\in S_2}\ 1 = \sum_{(a,x)\in S_1\times S_2} \ 1.
$$
Best Answer
For didactic reasons, I'll show you how you could salvage your definition. You had the right idea! I think it's good to get a feel for how salvaging something is done. That's what mathematicians do all day if a definition or theorem doesn't work out. After that, I'll show you how the current body of mathematics actually deals with the apparent non-commutativity of the cartesian product.
Adapting your definition
Let us first start with your "definition"
$$A \cdot B \cdot C = \{\{a,b,c\} : a\in A,b\in B, c\in C\}$$
Albeit you can define it as such, it will very much differ from the ordinary cartesian product. Consider $A = \{a\}, B = \{a, b\}, C= \{b, c\}$. Then according to your definition we have $$A \cdot B \cdot C = \{\{a, b\}, \{a, c\}, \{a, b, c\}\}.$$ Certainly, this does not represent the set of all possible choices we could make drawing from $A, B, C$ -- which would be what $A \times B \times C$ would represent. Especially, your definition only gives 2 such "choices", whereas we actually had $|A||B||C| = 1\cdot 2 \cdot 4$ choices with the usual cartesian product.
But your idea of using sets $\{\ldots\}$ instead of tuples to ensure commutativity was good! It threw away order on the elements, but also duplicates. The latter is where the definition broke down. So why not just use a mathematical "data structure" which allows duplicates, but ignores order? We can use multisets for that. They are like sets, but with them an element can occur more than one time. Let's denote them by square brackets $[\ldots]$ and use them:
$$A \cdot B \cdot C = [[a,b,c] : a\in A,b\in B, c\in C]$$
Accept for a second that I also used multisets on the outermost set. Let us recompute our example from above:
$$A \cdot B \cdot C = [[a, a, b], [a, a, c], [a, b, b], [a, b, c]]$$
Now let's test whether it's commutative:
$$B \cdot A \cdot C = [[a, a, b], [a, a, c], [b, a, b], [b, a, c]]$$
Indeed we have the same result! If you don't see it, note that we have multisets! In particular $[b, a, b] = [a, b, b]$. Everything is equal up to order.
Why did we need the outermost multiset as well? Consider $$A = \{a, c\}\\ B = \{b\}\\ C = \{a, c\}$$
With the usual cartesian product we would have the tuples $(a, b, c)$ and $(c, b, a)$ -- among others of course. With my definition above we would have $[a, b, c]$ and $[c, b, a]$, which are -- as multisets -- the same! So if we did not have the outer multiset, we would squash them to one single element. And that would again bring our cardinality of the supposed commutative cartesian product down -- the same problem as you had with your definition I described above.
What is actually used in mathematics
We usually stick to the usual cartesian product $A \times B = \{(a, b) : a \in A, b \in B\}$. To see why, consider the two notions of equality we have in many fields of mathematics:
In our case, we have $A \times B \neq B \times A$, but we have the second point fulfilled! $A \times B$ is isomorphic to $B \times A$. Do you see how?
And equality of things up to isomorphisms is good enough for many fields of mathematics. In fact, people tend to say the cartesian product is commutative because the existence of an isomorphism is good enough. (For a second, I wanted to answer your question with: it's commutative, what's your question again? :-))