quantum mechanics - 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 $P(q_1,q_2, \dots, q_n) = n! |\Psi (q_1,q_2, \dots, q_n)|^2$).