# Normalizing symmetric wavefunctions

Wikipedia and other sources say that the normalized symmetric ket for $$N$$ particles with quantum numbers $$n_1, n_2, ...,n_N$$ is $$|n_1n_2...n_N;S\rangle=\sqrt{\frac{\Pi_km_k!}{N!}}\sum_P|n_{P(1)}\rangle|n_{P(2)}\rangle...|n_{P(N)}\rangle$$ Where $$m_k$$ is the number of particles in stake $$k$$. My question is, since for every $$m$$ particles in a same state there are $$m!$$ permutations of the base kets, shouldn't the $$m_k!$$ be in the denominator to account for this?

Maybe the idea will be clear by a simple example. First consider $$|123 \rangle$$. The symmetrized version of this is proportional to $$|123\rangle + |132\rangle + |231\rangle + |213\rangle + |312\rangle + |321\rangle$$ with $$6 = 3!$$ entries in total. So we need to multiply by $$1/\sqrt{3!}$$ to normalize this state properly, in accordance with the formula. Now consider $$|112 \rangle$$. The symmetrized version is proportional to $$|112 \rangle + |121 \rangle + |211 \rangle.$$ Now there are only $$3 = 3!/2!$$ entries, so we need to multiply by $$\sqrt{2!}/\sqrt{3!}$$, again in agreement with the formula. In a more extreme case, consider $$|111 \rangle$$. The symmetrized version is just $$|111 \rangle$$, so the normalization constant is $$1 = \sqrt{3!}/\sqrt{3!}$$, also in agreement with the formula.
If you want a description in terms of words, what's going on is that our description should treat the particles symmetrically because they're identical. So if you have a state like $$|123 \rangle$$, you need to split it between $$N!$$ symmetric possibilities. That's what the $$1/\sqrt{N!}$$ factor account for. But if you have a situation where some of the particles are in the same state on the left-hand side, then those particles are already treated symmetrically. So there are fewer than $$N!$$ states on the right-hand side, which is why the $$\prod_k m_k!$$ appears in the numerator rather than the denominator.