# 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?

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*}