Varying Order of Covariant Differentiation Does the order of covariant differentiation matter?
Will   E$_i$$_j$,$_k$$_l$$_t$ =E$_i$$_j$,$_l$$_t$$_k$ ?
Does it matter if the tensor E$_i$$_j$ is continuously differentiable?
 A: In general no, for example on a reimannian manifold where the christoffel symbols and its derivatives do not vanish there is a loss in commutativity:
$$\nabla_a \nabla_bT^c \ne \nabla_b \nabla_aT^c$$
infact:
$$\nabla_a \nabla_bT^c - \nabla_b \nabla_aT^c = R^{c}_{dab}T^d$$
Where $R$ is called the reimann christoffel tensor. It has a beautiful story which you can find by searching for it in wikipedia along with a more geometric picture as a preamble to the concept of curvature.
In a euclidean space  this is not the case as the mixed partial derivatives are equal and the components of $R$ vanish. 
A: Yes, the order does matter, in just the way that the order of taking mixed partial derivatives matters, at least for functions whose partials exists but whose total derivative does not (a property usually due to non-commuting of the partials!). 
Take some such example -- which you can look up in most multivariable calculus books -- and you can build your tensor example. Call the example $x, y \mapsto f(x, y)$. 
Now build $E$: First, you can say that there's just one component, $E_{11}$, since $ij$ have no bearing on your question. Then make $E(x, y, z)$ be simply $z \cdot f(x, y)$. Look at 
$$
E_{x,y,z} \\
E_{y, z, x}
$$
The first will be $\frac{\partial^2 f}{\partial x \partial y}$, the second will be $\frac{\partial^2 f}{\partial y \partial x}$. These will then be distinct for some $(x, y)$, because $f$ is a function chosen to have unequal mixed partials at some point. 
If, on the other hand, everything in sight is infinitely continuously differentiable, then all such examples fly out the window, and reordering the derivatives will be fine. 
