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.           
 A: 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.
A: The number of ways in which 1, 2 and 3 can be arranged whereby order matters can be calculated using the formula $\frac{n!}{(n-r)!}$, where $n$ denotes the number of items that you can choose from, and $r$ denotes the number of items that you are choosing to arrange. This is called a permutation. In this case $n=r=3$, and so 1, 2 and 3 can be arranged in $\frac{3!}{0!}=\frac{(3)(2)(1)}{1}=6$ different ways.
The number of ways in which 1, 2 and 3 can be chosen whereby order does not matter can be calculated using the formula $\frac{n!}{(n-r)!r!}$, where $n$ denotes the number of items that you can choose from, and $r$ denotes the number of items that you are choosing. This is called a combination. In this case, again, $n=r=3$, and so 1, 2 and 3 can be chosen in $\frac{3!}{0!3!}=\frac{(3)(2)(1)}{(1)(3)(2)(1)}=1$ different way, whereby order does not matter.
A: If your'e asking a question like this I'm assuming you don't know any set theory (sorry if that's incorrect), so I'll stay informal.
What the answer is depends on the question. If I ask "how many ways can I get a straight hand in poker with a standard deck?", then order doesn't matter here. A poker hand is the same regardless of the order in which cards were drawn, so order doesn't matter for this question. In this case $(2,3,4,5,6)$ and $(3,4,6,2,5)$ are the same. They're both the same straight.
Let's take a completely different question: "I have 5 people running in a race. How many possible standings can I have at the end?" This question is all about order, since that's the only thing that' s changing here (we're considering all people the whole time). In this case, $(3,4,5,1,2)$ (3rd individual got 1st, 4th individual got 2nd, etc.) and $(2,3,4,5,1)$ are very different. 
Whether order matters or not is entirely based on the question. As others have explained, the mathematical objects we use to describe cases where order matters are called sequences, while those where order does not are called sets (and multisets, but let's not get into that).
