Words of length $n$ out of an alphabet of $n$ symbols such that no symbol appears exactly $k > 0$ times I need to count the possible words of length $n$ out of an alphabet of $n$ symbols such that no symbol appears exactly $k$ times, with $k > 0$.
I already tried considering the weak compositions of $n$ in $n$ parts with the restriction of having no parts being exactly $k$, but the problem is that then I would need to find a way of counting how many ways of ordering I have for each of the compositions.
I would need the exact amount or asymptotics or at least any bounds that are tighter than 
$$C^w_{n}(n,\check{k})n!$$
where $C^w_{n}(n,\check{k})$ is the amount of weak compositions of $n$ in $n$ parts with no parts being equal to $k$.
 A: As suggested in a comment, this can be solved using inclusion-exclusion.
There are
$$
\frac{n!}{(k!)^j(n-jk)!}(n-j)^{n-jk}
$$
words in which $j$ particular symbols appear exactly $k$ times. Thus by inclusion–exclusion there are
$$
\sum_{j=0}^{\left\lfloor\frac nk\right\rfloor}(-1)^j\binom nj\frac{n!}{(k!)^j(n-jk)!}(n-j)^{n-jk}
$$
words in which no symbol appears exactly $k$ times.
A: The combinatorial class of these words is given by
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=n}(\textsc{SET}_{\ne k}(\mathcal{Z})).$$
This gives the EGF
$$G(z) = \left(\exp(z)-\frac{z^k}{k!}\right)^n.$$
Extracting coefficients we find
$$n! [z^n] G(z) = n! [z^n]
\sum_{j=0}^n {n\choose j} \exp((n-j)z) (-1)^j \frac{z^{kj}}{(k!)^j}
\\ = n! [z^n]
\sum_{j=0}^{\lfloor n/k \rfloor}
{n\choose j} \exp((n-j)z) (-1)^j \frac{z^{kj}}{(k!)^j}
\\ = n!
\sum_{j=0}^{\lfloor n/k \rfloor}
{n\choose j} [z^{n-kj}] \exp((n-j)z) (-1)^j \frac{1}{(k!)^j}
\\ = n!
\sum_{j=0}^{\lfloor n/k \rfloor}
{n\choose j} \frac{(n-j)^{n-kj}}{(n-kj)!} (-1)^j \frac{1}{(k!)^j}.$$
This matches the result from PIE by @joriki.
