From Rotman's "Algebraic Topology":
Two orientations of $\Delta^n$ are the same if, as permutations of $\{e_0, \dots, e_n\}$, they have the same parity (i.e. both are even or both are odd); otherwise the orientations are opposite.
I don't understand this definition. What is the definition of the parity of an orientation of a simplex? What is the definition of an even and an odd orientation of a simplex?
If $n\geq 1$ an orientation on the standard $n$-simplex $\Delta^n$ is an ordering of its vertices modulo the group of alternating permutations $A_{n+1} \subset \Sigma_{n+1}$, where $\Sigma_{n+1}$ is the permutation group of the set of vertices $\{e_0,\dots,e_{n}\}$ with their standard ordering. The set of orientations on $\Delta^n$ is simply $\Sigma_{n+1}/A_{n+1}\cong \mathbb{Z}/2$, and the "parity" of an orientation is given by which $A_{n+1}$-coset it is in: if it is in $A_{n+1}$ it is "even" and if it is not then it is "odd". It is a result from group theory that a permutation is in $A_{n+1}$ iff it can be written as an even number of transpositions.
For a simple example, consider the $1$-simplex $\Delta^1$ which is the convex hull of $\{e_0,e_1\}$. The orientation $[e_0e_1]$ is even because it is represented by the identity permutation ($0$ transpositions), but the orientation $[e_1e_0]$ is odd because the premutation representing it is a transposition. This case is degenerate since $A_2$ is the trivial group, and $\Sigma_2\cong \Sigma_2/A_2$ is just $\{ (e_0 e_1), (e_1 e_0) \}$. (Here I'm using "$()$" to denote a permutation and "$[]$" to denote its $A_{n+1}$-coset, i.e. the orientation it determines.)
For $\Delta^2$ with vertices $\{e_0, e_1, e_2\}$, the even permutations are $(e_0 e_1 e_2)$, $(e_1 e_2 e_0)$, $(e_2 e_0 e_1)$, and the odd permutations are $(e_0 e_2 e_1)$, $(e_2 e_1 e_0)$ and $(e_1 e_0 e_2)$. As orientations, we have $[e_0 e_1 e_2]=[e_1 e_2 e_0] = [e_2 e_0 e_1]$ and $[e_0 e_2 e_1]=[e_2 e_1 e_0]=[e_1 e_0 e_2]$.