The symmetry normalization factor of wave functions in quantum statitics

Now I am studying statistical mechanics by R.K.Pathria & Paul D.Beale's Statistical Mechanics(3rd Edition). In page 134, the book claims that the wave function $\psi$ of a system composed of N indistinguishable, non interacting particles is $$\psi_{K}(1,...,N)=(N!)^{-1/2} \sum_{P} \delta_P P[u_{k_1}(1)...u_{k_N}(N)].$$ But as for boson in degenerate case, for example, if $k_1=k_2$, then there will be no permutation of 1 and 2 and the normalization factor is $\sqrt{\frac{2!}{N!}}$ rather than $\sqrt{\frac{1}{N!}}$. But if I take into account this, the following derivation will be too messy to get the density matrix of the system of free particles. So I want to know whether the approximation the book makes is reasonable or not and why. And the book claims that its derivation is still valid when the mean interparticle distance $(V/N)^{1/3}$ is much comparable to the mean thermal wavelength $\frac{h}{(2\pi mkT)^{1/2}}$ and the degenerate discriminant must be taken into account. Click link to see Pathria's discussion of the topic in his textbook.

• Why not represent states in terms of occupation numbers of single particle states? – Count Iblis Oct 24 '15 at 16:09
• Please describe what does your formula for the wave-function mean; what is $P,u$ and what is the space over which the summation is executed. I also do not understand what the notation $\psi(1,...,N)$ means, is this a single-particle wavefunction in an energy-level representation or some particular compact notation? – Void Oct 24 '15 at 20:34
• $u$ is the eigen wavefunction of energy operator of a single particle. $1,...,N$ are position variables $r_1,...,r_n$.P represents the permutation of the $r_1,...,r_n$ in the wavefuction. I have added a link to this section which describes all of them clearly. – Eric Yang Oct 25 '15 at 4:19