Derangements of multisets

The answer is given in this Wikipedia section: Up to a sign, the number of such derangements is given by

$$\int_0^\infty L_{n_1}(x)\cdots L_{n_r}(x)\mathrm e^{-x}\mathrm dx\;,$$

where $r$ is the number of items, $n_i$ is the number of occurrences of item $i$ and $L_k$ is the $k$-th Laguerre polynomial. For instance, for $1,1,2,2,14$, we have $r=3$, $n_1=2$, $n_2=2$, $n_3=1$, so up to a sign the number of derangements is

$$ \begin{align} \int_0^\infty L_2(x)L_2(x)L_1(x)\mathrm e^{-x}\mathrm dx &= \int_0^\infty \frac12(x^2-4x+2)\frac12(x^2-4x+2)(1-x)\mathrm e^{-x}\mathrm dx \\ &= \int_0^\infty\left(-\frac14x^5+\frac94x^4-7x^3+9 x^2-5x+1\right)\mathrm e^{-x}\mathrm dx \\ &=-4\;. \end{align} $$

This indicates an error in your example, and indeed, the arrangement $2,2,1,1,14$ is not a derangement since the $14$ is in the same place.