Simple Combinatorics Order does matter:  
$$\begin{align*}
&1 2 3\\
&1 3 2\\
&2 1 3\\                       
&2 3 1\\
&3 1 2\\
&3 2 1
\end{align*}$$
Order doesn't matter:
$$1 2 3$$
How does this work? Order doesn't matter: $1 2 3$. Can't it also be say, $3 2 1$? Doesn't this also show that order doesn't matter.           

The question isn’t very clear, but I think that you’re trying to say something like this:

If order matters, there are six permutations of the numbers $1,2$, and $3$: $123,132,213,231,312$, and $321$. If order doesn’t matter, there is just the set $\{1,2,3\}$. Can’t we also write that set $\{3,2,1\}$, for instance?

Yes: the set whose only members are the numbers $1,2$, and $3$ can be written in any of the following ways:
$$\begin{array}{}
\{1,2,3\}&\{1,3,2\}&\{2,1,3\}\\
\{2,3,1\}&\{3,1,2\}&\{3,2,1\}
\end{array}$$
Two sets are equal if and only if they have the same members; the order in which you list the members doesn’t matter. It doesn’t even matter if you list some members more than once: $$\{1,1,3,2,3,3,1,2,3,2,3,1\}=\{1,2,3\}\;,$$ though it’s hard to imagine why anyone would want to write it this way under normal circumstances. 
When we’re counting something and say that order doesn’t matter, we mean that we’re just counting sets of things. When order does matter, we’re counting permutations of things, like the six permutations of $1,2$, and $3$ that you listed. Informally you can think of a permutation as an ordered set, a set with a fixed order established for its elements; a plain old set has no such order.
