Normalization factor for symmetric state As per Wikipedia,
The above discussion generalizes readily to the case of $N$ particles. Suppose there are $N$ particles with quantum numbers $n_1, n_2, ..., n_N$. If the particles are bosons, they occupy a totally symmetric state, which is symmetric under the exchange of any two-particle labels:
$$|n_1,n_2,...,n_N;S\rangle =\sqrt{\frac{\prod_nm_n!}{N!}}\sum_p|n_{p(1)}\rangle|n_{p(2)}\rangle\cdots |n_{p(N)}\rangle $$
The quantity $m_n$ stands for the number of times each of the single-particle states $n$ appears in the N-particle state.

But Kardar's book, Statistical Physics of Particles, says
The bosonic subspace is constructed as
$$|\vec k_1,\cdots \vec k_N\rangle _+=\frac{1}{\sqrt{N_+}}\sum_P P |\vec k_1,\cdots \vec k_N\rangle_\otimes $$
Proper normalization requires $N_+=N! \prod_{\vec k}n_{\vec k}!$.
A particular one-particle state may be repeated $n_{\vec k}$ times in the list.
But both of them are quite different. Why it is so?
 A: I am also looking for an answer to this, but I believe it is the latter. I think about it like this:
There are definitely $\frac{N!}{\prod_{n}(m_n!)}$ distinct permutations that change $|n_1 \rangle \cdots |n_N \rangle$ which are orthonormal and for each of those distinct permutations, there are $\prod_{n}(m_n!)$ permutations that dont change $|n_1 \rangle \cdots |n_N \rangle$
For notational convenience define $k = \frac{N!}{\prod_{n}(m_n!)}$ and $l = \prod_{n}(m_n!)$
So we can group all permutations of the $|n_1 \rangle \cdots |n_N \rangle$ into $k$ groups, each group containing the same state $l$ times and picking a  representative state from each group, say states $\psi_1, \cdots, \psi_k$ gives us $k$ orthonormal states
So we can write
\begin{align*}
\bigg \| \sum_{\sigma \in S_N}   |n_{\sigma(1)} \rangle \cdots |n_{\sigma(N)}  \rangle \bigg\|^2
&= \sum_{i=1}^k \| l \psi_i  \|^2 \\
&= l^2 \sum_{i=1}^k \| \psi_i\|^2 \\
&= l^2 k \\
&= N! \prod_{n}(m_n!) 
\end{align*}
Let me know what you think.
