Let $S$ be a finite set of primes and $K/\mathbb{Q}$ a finite Galois extension unramified outside $S$. For each primes $p\notin S$, let $n_p$ be the number of primes of $K$ above $p$ and $f_p$ the degree of the residue field extension. An elementary result in algebraic number theory says,
$n_pf_p=[K:\mathbb{Q}]$.
With the setting, my question is: for any pair of integer $(n,f)$ such that $nf=[K:\mathbb{Q}]$, does there exist $p\notin S$ such that $n_p=n$ and $f_p=f$?
The answer is no. Consider $ K = \mathbf Q(\sqrt{2}, \sqrt{3}) $, and let $ p $ be any prime unramified in this extension. We will show that $ p $ cannot be inert. Assume the contrary, and let $ G $ be the decomposition group of $ p $, which is necessarily the whole Galois group. There is then a surjection from $ G $ to the Galois group of the residue class field extension. This extension is the quartic extension of $ \mathbb F_p $, so its Galois group over $ \mathbb F_p $ is $ C_4 $. But then, we must have a surjection $ G \cong C_2 \times C_2 \to C_4 $. Such a surjection would be an isomorphism, which is absurd. Therefore, no prime of $ \mathbf Z $ is inert in $ K $, thus the equation $ n_p f_p = [K : \mathbf Q] $ has no solution for $ f_p = [K : \mathbf Q] $ and $ n_p = 1 $. Similar counterexamples are easily constructed using the same idea.