Unordered cartesian product? I have a set $\Omega=\{1;2;6\}$ and I want to define another set $A$ consisting of all triples $(a,b,c)$ with $a,b,c\in\Omega$, which contain exactly two 6's.
My first attempt looked like this: $A=\{a^2\times b\vert a=\{6\},b=\{1;2\}\}$
But after looking at the cartesian product's definition, it looks like my set $A$ consists of all triples having two 6's as their first two elements instead of all triples containing two 6's. So here is my second attempt: $A=\{(a^2\times b)\cup(a\times b\times a)\cup(b\times a^2)\vert a=\{6\},b=\{1;2\}\}$
So my questions are, if my last definition of $A$ would be correct (?), if there is a way to shorten the definition of $A$ (?) and if there is something like an "unordered cartesian product"?
 A: There is an "unordered Cartesian product." Given a set $X$, we define $S^k(X)$ to be the quotient of $X\times X\times\cdots \times X$ by the action of the symmetric group on $k$ letters. This is sometimes called the $k$th symmetric power.
A: The simplest way that does not involve permutations or functions is to directly express what you want:
  $\{ (a,b,c) : (a,b) = (6,6) \lor (b,c) = (6,6) \lor (c,a) = (6,6) \}$
A slightly shorter alternative based on the fact that $\{6,6\} = \{6\}$ is:
  $\{ (a,b,c) : \{a,b\} = \{6\} \lor \{b,c\} = \{6\} \lor \{c,a\} = \{6\} \}$
A: Your idea betrays a minor discomfort with the notation. Literally, $\{(a^2\times b)\cup (a\times b\times a)\cup (b\times a^2)|a=\{6\},b=\{1;2\}\}$ is a set with a single element, namely the set $B=(a^2\times b)\cup (a\times b\times a)\cup (b\times a^2)$, where $a$ is a set with one element, namely $6$, and $b$ is a set with two elements, namely $1$ and $2$. $B$, however, is exactly the set you want. 
The notation you choose is more characteristic when $a$ and $b$ are varying elements of a set, rather than fixed sets themselves. For example we could refer to the natural numbers as $\mathbb{N}=\{n| n\geq 0, n\in \mathbb{Z}\}$, where here $n$ actually varies, namely over the integers.
