Let nonnegative integer decision variable $x_k$ represent the number of $k$-gons. A necessary condition is that the internal angles sum to $360$ degrees:
$$\sum_{k \ge 3} 180 \frac{k-2}{k} x_k = 360. \tag0\label0$$
I found
>! 17 solutions to \eqref{0}:

\begin{align}
&(x_{6}=3 )\tag{1}\label{1}\\
&(x_{5}=2 ,x_{10}=1 )\tag{2}\label{2}\\
&(x_{4}=1 ,x_{8}=2 )\tag{3}\label{3}\\
&(x_{4}=1 ,x_{6}=1 ,x_{12}=1 )\tag{4}\label{4}\\
&(x_{4}=1 ,x_{5}=1 ,x_{20}=1 )\tag{5}\label{5}\\
&(x_{4}=4 )\tag{6}\label{6}\\
&(x_{3}=1 ,x_{12}=2 )\tag{7}\label{7}\\
&(x_{3}=1 ,x_{10}=1 ,x_{15}=1 )\tag{8}\label{8}\\
&(x_{3}=1 ,x_{9}=1 ,x_{18}=1 )\tag{9}\label{9}\\
&(x_{3}=1 ,x_{8}=1 ,x_{24}=1 )\tag{10}\label{10}\\
&(x_{3}=1 ,x_{7}=1 ,x_{42}=1 )\tag{11}\label{11}\\
&(x_{3}=1 ,x_{4}=2 ,x_{6}=1 )\tag{12}\label{12}\\
&(x_{3}=2 ,x_{6}=2 )\tag{13}\label{13}\\
&(x_{3}=2 ,x_{4}=1 ,x_{12}=1 )\tag{14}\label{14}\\
&(x_{3}=3 ,x_{4}=2 )\tag{15}\label{15}\\
&(x_{3}=4 ,x_{6}=1 )\tag{16}\label{16}\\
&(x_{3}=6 )\tag{17}\label{17}\\
\end{align}

(Solution \eqref{4} corresponds to the example given in the question.)

Up to rotation and reflection, solutions \eqref{12} through \eqref{15} yield two arrangements, and the rest yield only one.  So in total there are
>! $17+4=21$ arrangements.