Normalising multi-particle wavefunctions For a quantum mechanical system of $n$ particles the state of the system is given by a wave function $\Psi (q_1, \dots , q_n)$. If the particles are indistinguishable, we demand that the swapping of two particles preserves the modulus of $\Psi$.
Suppose we want to work out the probability density of finding the particles in the positions $(Q_1, \dots Q_n)$. This should just be $|\Psi (Q_1, \dots Q_n)|^2$. But if permuting the particles amongst these positions represents the same event, then the normalisation condition should be $$\int |\Psi (q_1, \dots q_n)|^2 d^n q = n!$$ rather than 1. That is we overcount by a factor of $n!$ because the different permutations are really the same event. Is this correct? Or is the correct probability density $n! |\Psi (Q_1, \dots Q_n)|^2 $? This makes sense to me but I'm unsure because I have never seen it stated in any textbook on quantum mechanics.
EDIT: I want to make clear exactly what I see the problem as being. Let's simplify and assume two particles and assume position is discretised so that there are two possible positions ($q=1$ or $q=2$).
Let the wave function be $\Psi (q_1, q_2)$. Normalisation says that
$$\sum_{q_1=1}^2 \sum_{q_2=1}^2 |\Psi (q_1, q_2)|^2 = 1 = |\Psi (1, 1)|^2 + |\Psi (1, 2)|^2 + |\Psi (2, 1)|^2 + |\Psi (2, 2)|^2$$
Notice there are four terms here. But if we think of particles as being indistinguishable there are only three possible outcomes: Both particles in position 1, both particles in position 2, one in each position. In other words, the events (1,2) and (2,1) are the same event. But when we normalise we necessarily double-count.
The principle of indistinguishability says that $|\Psi (1, 2)|^2 = |\Psi (2, 1)|^2$. But these cannot be the probabilities. Normalisation of probability says that $$P(1,1) + P(1,2) + P(2,2) = 1$$ But if $P(i,j) = |\Psi (i, j)|^2$, the normalisation condition gives $P(1,1) + 2P(1,2) + P(2,2) = 1$ which is a contradiction. As I see it, the solution is that $P(1,2) = 2 |\Psi (1, 2)|^2$ (or more generally $P(q_1,q_2, \dots, q_n) = n! |\Psi (q_1,q_2, \dots, q_n)|^2$).
 A: No, the normalization condition is always the same. It has to be so from the way averages of observables are defined:
$$
\langle A\rangle|_\Psi:=\frac{\langle\Psi|A|\Psi\rangle}{\langle\Psi|\Psi\rangle}.
$$
States are taken to be normalized in order to avoid the denominator. It's just a comfortable choice, and there is no reason to change it. When you refer to the wavefunction $\Psi(q_1,...q_N)$ for a system of $N$ identical particles (bosons or fermions), you are taking a normalized wavefunction.
Your confusion in the case of identical particles can be addressed by an explicit example, that is very often found in practical applications. Suppose that we have $N$ wavefunctions $\psi_i$ that take as input only one coordinate (or set of three coordinates, in 3D), and we want to describe a set of $N$ fermions by assigning to each one a wavefunction. We take each of the wavefunction as a distinct wavefunction, representing a distinct state (this is the most general case, as we're dealing with fermions), and we impose orthonormality relations $\langle \psi_i|\psi_j\rangle=\delta_{ij}$.
After this preamble, we want to build a wavefunction for the $N$ fermions. We start with a tentative wavefunction like
$$
\Psi(q_1,...,q_N)=\psi_1(q_1)...\psi_N(q_N).
$$
This wavefunction has unit norm, due to the orthonormality of the $\psi_i$, but is not antisymmetric with respect to the exchange of fermion coordinates $q_i\leftrightarrow q_j$. It is interpreted as "the first fermion is in the state described by $\psi_1$, the second in the state described by $\psi_2$ et cetera", so it is clearly making distinctions between particles.
To solve this problem, we antisymmetrize the wavefunction as
$$
\Psi(q_1,...,q_N)=\frac{1}{\sqrt{N!}}\sum_{P}(-1)^{P}\psi_{P(1)}(q_1)...\psi_{P(N)}(q_N).
$$
This notation means "sum over all permutations $P$ of the string $1...N$, leaving the coordinates fixed and moving the wavefunction indices, and inserting a sign $-$ for odd permutations" (you can also do it the other way round). Now, here you have a $\sqrt{N!}$ like the one that is bothering you. This is exactly what is needed to normalize the entire wavefunction. Why?
Well, when you perform the integral of $\Psi^*\Psi$, it is actually a sum of products of single particle integrals, of the form $\int\psi_i^*(q)\psi_j(q)dq$, and the orthonormality condition ensures that, for the integral to be zero, you must have the same wavefunctions under the integral $(i=j)$. In the modulus of $\Psi$ the only surviving contributions are then the integrals where a permutation of $(1...N)$ in the bra is exactly matched by the same permutation in the ket. Any of those terms gives $1$, due to orthonormality. Now, how many permutations of a string of $(1...N)$ can you make? Exactly the $N!$ that is needed to cancel the denominator.
A: When (anti)symmetrizing wavefunctions of a system of identical particles, one has restricted the state vectors to live in a subspace of the original Hilbert space, and thus the orthonormality and completeness relations become different. Your question in fact involves both of these relations.
First, one recalls that in QM, if two state vectors are proportional to each other (two vectors align), they are said to be equivalent, and only one of them contributes in the completeness relation. In many-body system, if one "permutates" two particles, the new state vector is equivalent to the original one. Consequently, one has
$$\sum_{q_1,q_2,\cdots,q_N}\frac{\prod_{q_j}n_{q_j}!}{N!}\left|q_1,q_2,\cdots,q_N\right\rangle\left\langle q_1,q_2,\cdots,q_N\right| = 1.$$
Here, $N$ is the total number of particles and $n_{q_i}$ the occupation number of single-particle state $q_i$. This coefficient eliminates the "overcount" of equivalent states: since there are $N$ particles, there are $N!$ permutations of them and thus one divides it by $N!$ when summing. However, in case of bosonic system, if there are $n_{q_i}$ particles occupying state $q_i$ then one has $n_{q_i}!$ same states (instead of equivalent), which means that number of equivalent states is reduced to $\dfrac{N!}{n_{q_i}!}$, etc. In continuous case, for example $q_i=x_i$,
$$\int\frac{dx_1dx_2\cdots dx_N}{N!}\left|x_1,x_2,\cdots,x_N\right\rangle\left\langle x_1,x_2,\cdots,x_N\right| = 1.$$
Here $n_{q_i}!=1$ since a position $x_i$ can only accommodate 0 or 1 particle. Combining this relation with the normalization relation one obtains your formula:
$$1=\left\langle q_1,q_2,\cdots,q_N|q_1,q_2,\cdots,q_N\right\rangle = \int\frac{dx_1dx_2\cdots dx_N}{N!}|\left\langle x_1,x_2,\cdots,x_N|q_1,q_2,\cdots,q_N\right\rangle|^2.$$
Hence, $|\left\langle x_1,x_2,\cdots,x_N|q_1,q_2,\cdots,q_N\right\rangle|^2$ is the probablity density for $N$ particles to be in states $q_1$, $q_2$, etc. On the other hand, the many-body state vector can be written in terms of single-particle states as
$$\left|q_1,q_2,\cdots,q_N\right\rangle = \frac{1}{\sqrt{N!\prod_in_{q_i}!}}\sum_{\mathcal{P}}\zeta^P\left|q_{P1}\right\rangle\otimes\left|q_{P2}\right\rangle\otimes\cdots\left|q_{PN}\right\rangle .$$
Its coordinate representation is
$$\left\langle x_1,x_2,\cdots,x_N|q_1,q_2,\cdots,q_N\right\rangle = \sqrt{N!}\Psi_{q_1,q_2\cdots,q_N}(x_1,x_2,\cdots,x_N).$$
Notice that $\left|r_1,r_2,\cdots,r_N\right\rangle = \dfrac{1}{\sqrt{N!}}\sum_{\mathcal{P}}\zeta^P\left|r_{P1}\right\rangle\otimes\left|r_{P2}\right\rangle\otimes\cdots\left|r_{PN}\right\rangle$ and $\Psi_{q_1,q_2\cdots,q_N}(x_1,x_2,\cdots,x_N)$ is the familiar many-body wavefunction one sees in textbooks, which is normalized to 1. In this case, $|\Psi_{q_1,q_2\cdots,q_N}(x_1,x_2,\cdots,x_N)|^2$ is the probability density that particle 1 at state $q_1$, particle 2 at state $q_2$, etc. Its physical meaning is different from the previous one.
Apply these general expressions into your two-particle example, you will see it more clearly.
For detailed derivation for some of these formulas, refer to Statistical mechanics: A set of lectures - Feynmen (chapter 6, section 6.7). To gain a brief insight into the difference between $|\Psi_{q_1,q_2\cdots,q_N}(x_1,x_2,\cdots,x_N)|^2$ and $|\left\langle x_1,x_2,\cdots,x_N|q_1,q_2,\cdots,q_N\right\rangle|^2$, refer to Principles of quantum mechanics - Shankar (Chapter 10).
