# Are tensored qubits commutative?

I am given a solution to a problem, saying that

$$|\psi\rangle_{ABC} = \frac{1}{\sqrt{2}}(|0\rangle_A \otimes |1\rangle_C + |1\rangle_A \otimes |0\rangle_C) \otimes |+\rangle_B$$

$$= \frac{1}{2}(|001\rangle_{ABC} + |100\rangle_{ABC} + |011\rangle_{ABC} + |110\rangle_{ABC})$$

However, simply performing out the tensors would give the same results but with the second and third qubits reversed, i.e.

$$\frac{1}{2}(|010\rangle_{ACB} + |100\rangle_{ACB} + |011\rangle_{ACB} + |101\rangle_{ACB})$$

since $B$ is on the right. Am I mistaken thinking that qubits cannot commute like that? How is the given answer correct?

It may be clearer if instead of all qubits, you put $A$ as a qudit with $d=d_A$ etc for B,C.
The first state is then in the Hilbert space $\mathbb{C}^{d_A} \otimes \mathbb{C}^{d_B} \otimes \mathbb{C}^{d_C}$
but the second is in $\mathbb{C}^{d_A} \otimes \mathbb{C}^{d_C} \otimes \mathbb{C}^{d_B}$
Now specialize to $d_A=d_B=d_C=2$. Now you should be more clear about which $2$ went where.