For a vector space $V$ and a linear operator $R:V \otimes V \to V \otimes V$, we say that $R$ satisfies the **Yang--Baxter equation** if
$$(R\otimes id)(id\otimes R)(R\otimes id)
 = (id\otimes R)(R\otimes id)(id\otimes R).$$
If instead $R$ satisfies
$$R\_{12}R\_{13}R\_{23} = R\_{23}R\_{13}R\_{12}$$
we say that $R$ satisfies the **quantum Yang--Baxter equation**.


So what is the difference between the Yang--Baxter equation and the quantum Yang--Baxter equation? I guess that YBE came first and then came QYBE, but I don't see what is quantum about QYBE. What different properties do both have, and why should one consider them two versions of the same thing?



What is the difference between the Yang--Baxter equation and the quantum Yang--Baxter equation?
Your two equations are equivalent, and are both versions of the quantum YBE. (The [question](https://math.stackexchange.com/questions/29054/what-are-the-differences-between-classical-yang-baxter-equation-and-quantum-yang) from the comments does a good job of answering your classical versus quantum question.)
Write the first as
$$
(c \otimes id)(id \otimes c)(c \otimes id) = (id \otimes c)(c \otimes id)(c \otimes id)
$$
and the second as
$$
R\_{12}R\_{13}R\_{23} = R\_{23}R\_{13}R\_{12}.
$$
Let $\tau : V \otimes V \to V \otimes V$ be the permutation map $\tau(v\_1 \otimes v\_2) = v\_2 \otimes v\_1$.
Then a solution $R$ to the second gives a solution $c = \tau R$ to the first, and vice-versa.
Whether or not to include the permutation in the definition of $R$-matrix is just a convention.


In the quantum groups literature, $V$ is a module for a quasitriangular Hopf algebra $H$ and $R$ can be given by the action of the *universal $R$-matrix* $\mathcal{R} \in H \otimes H$. Since $\tau$ will not be given by the action of something in $H \otimes H$, it's natural to leave it out.
(I think the YBE for $R$ is also more natural in the spin chain context, but I know less about that.)


However, in quantum topology we focus more on $c = \tau R$ (because it gives braid group representations) and less on $R$, so some authors eliminate the distinction and call $c$ an $R$-matrix. I prefer to call $c$ a *braiding* and $R$ an *$R$-matrix*, but this convention is not universal.


