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When using the cartesian product, and you have three collections, do you take two collections at a time OR all three together to calculate the product?

My question is if you have more than two collections, let's say A, B and C

A = {1,2,3} B = {4,5,6} C = {7,8}  A x B x C {1,2,3} x {4,5,6} x {7,8} 

Do you with the cartesian product calculate A x B, then B x C? And maybe A x C? Which means you take only two collections at a time.

OR

Do you take all three collections at the same time A x B x C?

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    In the usual formal buildup, it is in principle two at a time. However, let $I_n$ be a fixed $n$-element set, say the set $\{1,2, \dots,n\}$. We could define the $n$-fold Cartesian product $A_1\times\cdots\times A_n$ as the set of all functions $f$ from $I_n$ to $\cup A_i$ such that $f(i)\in A_i$ for $i=1$ to $n$.2012-01-30

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For $n \in \mathbb{N}$, the $n$-ary Cartesian product of $n$ sets $A_1, \dots, A_n$, denoted $A_1 \times \cdots \times A_n$, is defined to be the set of all $n$-tuples $(a_1, \dots, a_n)$ for which $a_i \in A_i$ for each $i$.

So in particular

$A \times B \times C = \{ (a,b,c)\, :\, a \in A,\ b \in B,\ c \in C \}$

This is distinct from

$(A \times B) \times C = \{ ((a,b),c)\, :\, a \in A,\ b \in B,\ c \in C \}$

each of whose elements is an ordered pair, the first 'coordinate' of which is itself an ordered pair.

Nonetheless, there is a very natural bijection

$\begin{align} A \times B \times C & \to (A \times B) \times C \\ (a,b,c) &\mapsto ((a,b),c) \end{align}$

and similarly for $A \times (B \times C)$.

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The cartesian product is an operation defined on two sets. Given the sets A and B the product A x B is not equal to the product B x A. So you will have to use two sets at a time and you will need to define an order, you want to apply the operation in, since (A x B) x C is not equal to A x (B x C).

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all three sets at the same time.