Suppose you have $N$ particles, each of which can occupy any of $s$ states. In general, you can write the $N$ particle Hilbert space $\mathcal{H}^N$ as a product of $1$ particle Hilbert spaces $\mathcal{H}^1$: $$ \mathcal{H}^N = \mathcal{H}^1 \otimes \mathcal{H}^1 \otimes \dots \otimes \mathcal{H}^1, $$ with $\mathrm{dim}[\mathcal{H}^1]=s$.
This means that the $N$ particle space will have $\mathrm{dim}[\mathcal{H}^N]=s^N$.
Now we look at the usual subspaces: $\mathcal{F}^N$ for fermions and $\mathcal{B}^N$ for bosons. For their dimensions, we have $$ \mathrm{dim}[\mathcal{F}^N] = \binom{s}{N} $$ and $$ \mathrm{dim}[\mathcal{B}^N] = \binom{s+N-1}{N}. $$ Now, I was under the impression that the symmetrization postulate, saying that there are only either completely symmetric or completely antisymmetric states, means that there is a decomposition of $\mathcal{H}^N$ into a direct sum $$ \mathcal{H}^N = \mathcal{F}^N\oplus\mathcal{B}^N. $$ However, as one can easily check (e.g. for $N=3$), this cannot be true since the dimensions don't add up, $\mathrm{dim}[\mathcal{H}^3] = g^3 \neq \mathrm{dim}[\mathcal{B}^N] + \mathrm{dim}[\mathcal{F}^N] \approx \frac13 g^3$.
What happens with the "missing" dimensions? Can something be said about a decomposition of $\mathcal{H}^N$ as a consequence of the symmetrization postulate?