$\randfa_{N}$, we have lost all feature labels. So $\randfa_{N}$ is a feature allocation, not an a priori labeled feature allocation.

Then the probability of the feature allocation $\randfa_{5} = \detfa_{5} := \{\{2,3\},\{2,3\}\}$ is $$q_{A}^{2} (1-q_{A})^{3} q_{B}^{2} (1-q_{B})^{3},$$ but the probability of the feature allocation $\randfa_{5} = \detfa'_{5} := \{\{2,3\},\{2,5\}\}$ is $$2 q_{A}^{2} (1-q_{A})^{3} q_{B}^{2} (1-q_{B})^{3}.$$ The difference is that in the latter case the features can be distinguished, and so we must account for the two possible pairings of features to frequencies $\{q_{A},q_{B}\}$.

Now, instead, let $\labrandfa_{N}$ be $\randfa_{N}$ with the features ordered uniformly at random amongst all possible feature orderings. There is just a single possible ordering of $\detfa_{5}$, so the probability of $\labrandfa_{5} = \labdetfa_{5} := (\{2,3\}, \{2,3\})$ is again $$q_{A}^{2} (1-q_{A})^{3} q_{B}^{2} (1-q_{B})^{3}.$$ However, there are two orderings of $\detfa'_{5}$, each of which is equally likely. The probability of $\labrandfa_{N} = \labdetfa'_{5} := (\{2,5\}, \{2,3\})$ is $$q_{A}^{2} (1-q_{A})^{3} q_{B}^{2} (1-q_{B})^{3}.$$ The same holds for the other ordering.

This example suggests that there are combinatorial factors that must be taken into account when working with the distribution of $\randfa_{N}$ directly. The example also suggests that we can avoid the need to specify such factors by instead working with 