Planes of a dodecahedron divide the space $\mathbb R^3$ In how many disjunct parts divide the planes of a regular dodecaheadron the space $\mathbb R^3$?
I encountered this task at university where we discussed it a lot, but nobody found a solution.

Does anybody know the solution? May it work with graph theory?
Here's a graphical count, based on intersecting the configuration with spheres of different sizes -- though in order to make things visible actually I'm normalizing the size of the spheres, such that the images here actually shows a sphere intersected by the planes of a dodecahedron of increasing size.

The 12 planes of the dodecahedron come in 6 parallel pairs, and in the figures, the regions of each sphere are color-coded according to (and labeled with) the number of pairs where the region is between the two parallel planes.




The first image shows a sphere far from the dodecahedron. Each of the 6 pairs of planes become a narrow band around the sphere, and the bands create the graph of an icosidodecahedron. Each face, edge, and vertex of the icosidodecahedron corresponds to an unbounded region of space -- there are $12$ of type 0, $20$ of type 0', $60$ of type 1, and $30$ of type 2.

As we zoom in towards the dodecahedron (or the dodecahedron become larger), the edges of each band move away from each other. The first change that happens is that the the 0' regions disappear and are replaced by $20$ cells of type 3. These new cells are the triangular spires of the great stellated dodecahedron.




Nothing new happens in the first of these pictures, except that the bands keep growing wider.

In the second picture the bands have just met in groups of five, and the last of the 0 and 1 regions disappear. Instead we get cells that are between 4 and 5 pairs of parallel planes.




The new yellow regions of type 4 merge, and we can see there are $30$ such cells, just as there were thirty 2s before, corresponding to the edges of either an icosahedron or a dodecahedron. In fact each of these cells is an irregular tetrahedron that shares one edge with the central dodecahedron, and the opposite edge with a great dodecahedron (whose edge figure looks like an icosahedron).

Removing the yellow tedrahedra from the great dodecahedron leaves a small stellated dodecahedron, whose pentagonal spires are the $12$ magenta 5 cells that came into being together with the 4s.




Finally a single red 6 cell appears: the original dodecahedron in the middle.

So how many regions were that? Summing everything we get
$$ \underbrace{(12+20)+60+30}_{\text{unbounded}}+\underbrace{20+30+12+1}_{\text{bounded}} = 185 $$
I found 432 in the book
"Les nombres remarquables" by Francois Le Lionnais (Herman, 1983)
(which is a list of interesting numbers), but there is no reference or proof. Among these 432 parts 192 are bounded.