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?
 A: 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. 
