What is the significance of the order in an ordered basis/basis? Throughout my Linear Algebra course I heard reference to the fact that a set must be ordered in some way to be a basis for a space, but never managed to see the importance of this - what is it? What would the consequences be of re-ordering our basis (apart from having to write matrix/vector representations in the some permuted form)?
The motivation of this question is that I've come to a problem where I am to find the dual bases corresponding to a certain bases of some space V. In doing so I've had to take care to order the dual bases carefully with respect to the respective bases of V, which made me wonder where else that the order of a basis becomes significant.
 A: As the others have pointed out, one fundamental use of ordered basis is to specify coordinates. Often, when we say something like $v=(1,2,3)$, we have already assumed that we are talking about coordinates w.r.t. an ordered basis: the canonical basis. If the order is not known, all we know is that $v$ has three coordinates $1,2,3$, but which one is the $x$-coordinate and which ones are $y$ and $z$ would be unknown to us, and hence we cannot be sure where exactly the point $v$ is.
Another purpose of ordering a basis is to make the matrix representation of a linear operator simpler. For instance, consider $f:\mathbb{R}^5\to\mathbb{R}^5$, where the matrix representation of $f$ w.r.t. the canonical basis $A=\{e_1,\ldots,e_5\}$ is given by
$$
[f]_A^A=\begin{pmatrix}
3&0&4&0&2\\
5&4&0&0&4\\
0&0&2&0&1\\
2&0&5&1&3\\
0&0&0&0&1
\end{pmatrix}.
$$
Can you tell me the value of $\det f$? For some, the answer can be obtained using only mental calculation, but for the others, the answer is not that easy to obtain. However, if we reorder the basis as $B=\{e_4,e_2,e_1,e_3,e_5\}$, the answer will become utterly obvious:
$$
[f]_B^B=\begin{pmatrix}
1&0&2&5&3\\
0&4&5&0&4\\
0&0&3&4&2\\
0&0&0&2&1\\
0&0&0&0&1
\end{pmatrix}.
$$
Admittedly, this example is a bit contrived, but the reordering of bases does sometimes help simplifying the structure of a matrix. For instance, in my answer to a recent question, I had used this trick to make the matrix representation of a certain linear operator block upper triangular, so that I could write down the determinant of this operator immediately.
A: As you've probably inferred from previous answers, the need to order a basis comes primarily from the desire to have the components of vectors (of coordinates), the rows of matrices, and the columns of matrices ordered.  That desire is a practical necessity if the vectors and matrices are to be written in the usual format on the page.  But if you're willing to think more abstractly  of the component positions in a vector, the row-labels, and the column-labels of a vector as being given by some index set $I$ other than the traditional $\{1,2,\dots d\}$, then you can perfectly well use $I$ to index (rather than order) your basis vectors, as Hagen von Eitzen suggested.  Furthermore, even this indexing isn't really necessary.  Given a basis, just as a set of vectors, you could use the basis vectors themselves as labels for the components of vectors and the rows and columns of matrices.  (This amounts to using the basis as your index set and indexing each basis vector by itself.)  
A: Think about the coordinates of a vector in your space relative to a certain ordered basis. They obviously change when you re-order your basis.
A: Even more, one should avoid using sets when talking about basis (or linearly independent systems). Otherwise one has confusing effects.
For example the set of column vectors of the matrix
$$\left(\begin{matrix}1&2&3\\4&5&5\\4&5&5\end{matrix}\right)$$
is linearly dependent, whereas the set of row vectors is linearly independent!
The preferred way is to speak of a family of vectors, $\{v_i\}_{i\in I}$, indexed by some index set $I$. 
It is the index set $I$ that enforces "order", for example if $I=\{1,2\}$, then there is a first vector $v_1$ and a second vector $v_2$ and swapping these makes a different basis.
A: If you study the problem of Linear Transformation given below I think you will know the significance of ordered basis.

A linear mapping  $T: R^3 \to R^2$ is defined by $T(x_1,x_2,x_3)=(3x_1-2x_2+x_3,x_1-3x_2-2x_3)$.Find the matrix of $T$ relative to the ordered basis $((1,0,0),(0,1,0),(0,0,1))$ of $R^3$ and $((1,0,0),(0,1,0))$ of $R^2$.

In this problem if you break the order of the basis then you must not get the desired result.
That's all from me.
A: Just as your coordinates are ordered and $(x,y)$ is to be distinguished from $(y,x)$, so should your basis. Because you want to use them in construction of other vectors. As in saying mix two of the first basis vector with three of the second basis vector to get some vector $v=2e_1+3e_2$, which is different from whatever $3e_1+2e_2$ is.
A: Another perspective: you use a basis as a way to represent vectors.  Vectors are geometric quantities that exist regardless of a basis.  But to describe vectors, it is convenient to use coordinates with respect to some fixed basis.  So I would say that you don't necessarily need an order for your basis, so much as an indexing set (which you can choose to linearly order if you so desire).  However, if you want to write your vector as a column vector or uses matrices to represent linear transformations, then they come with a natural ordering and this will force you to order your basis. 
A: What Dmitry said is correct.In addition, certain binary operations like the cross product depends on order.
