Is bosonic state space a proper subspace of the original Hilbert space on which we have an $N$-particle free Hamiltonian?

This question may be pretty naive. But I wanted to confirm it anyway. Also if people can tell me more about this fact, I will be grateful.

I am looking at a simple problem of n-particle free Hamiltonian:

begin{align}
label{1}
tag{1}
H_N=sumlimits_{i=1}^NH_i ;with; H_i=-frac{hbar^2}{2m}nabla^2_{vec{r}_i}+V(vec{r}_i)
end{align}

Now, suppose the single particle energy eigenstates are ${u_j(vec{r}_i)}_{j=0}^{n}$ such that begin{align}tag{2}label{2}H_iu_j(vec{r}_i)=e_ju_j(vec{r}_i)end{align}

We can very easily list a basis for the Hilbert space spanned by the eigenstates of $H_N$, viz. $$tag{3}label{3}{prodlimits_{i=1}^Nu_{j_{i}}(vec{r}_i):j_i text{ picks out an energy eigenstate of $H_i$ }}$$ and each such eigenstate in the list given has an eigenvalue of $sumlimits_{i=1}^{N}e_{j_i}$.

Now, I want to consider the bosonic state space only. Then, I take the basis elements from (ref{3}) and construct all possible sum of permutations of the form:
begin{align} tag{4}label{4}{sumlimits_{text{permutations of} (p_{1},p_{2},…,p_{k})}prodlimits_{i=1}^Nu_{j_i}(vec{r}_i):j_iin {p_1,p_2,…,p_k}, {p_1,p_2,…,p_k} text{ label a set of k distinct energy eigenstates of } H_i}
end{align}

Thus the new basis,(ref{4}), consisting of only symmetrized eigenstates is smaller in size than the one in (ref{3}).

Is this correct? For then it would mean the bosonic sector of the starting Hilbert space is a proper subspace of it. Same would hold for the fermionic sector.

Is the combination of bosonic and fermionic sector also a proper subspace of the starting Hilbert space?

Soliciting comments and answers.