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"?
Best Answer
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.