Does $\mathbf{n}\otimes \mathbf{m}=\mathbf{m}\otimes \mathbf{n}$ in the Clebsch-Gordan decomposition?

Question

Consider the two irreducible representations, $\mathbf{n}$ and $\mathbf{m}$, of $su_\Bbb{C}(N)$. The tensor product $\mathbf{n}\otimes \mathbf{m}$ can be found using Young tableaux as follows:

1. Write $\mathbf{n}$ and $\mathbf{m}$ as Young tables and fill the second one (i.e. $\mathbf{m}$) with the letters $a$ in the first row, $b$ in the second and so on.

2. Attach boxes from $\mathbf{m}$ to $\mathbf{n}$ such that:

   1. We have a Young tabulaux
   2. When reading the numbers from right to left and top to bottom the number of $a$s must exceed or be equal to the number of $b$s which must exceed or equal the number of $c$s etc. (i.e. a lattice permutation).

My question is: Instead of filling the boxes of $\mathbf{m}$ with letters and adding them to $\mathbf{n}$ can we fill the boxes of $\mathbf{n}$ with letters and add them to $\mathbf{m}$? i.e. in the context of Clebsch-Gordan decompositions does $\mathbf{n}\otimes \mathbf{m}=\mathbf{m}\otimes \mathbf{n}$