Covering the hypericosahedron with independent cells

The 600-cell is the finite regular four-dimensional polytope with Schläfli symbol {3, 3, 5}. It is also called hypericosahedron, because it can be regarded as the 4-dimensional analog of the icosahedron. The 600-cell is composed of 600 tetrahedra, with 5 to an edge. It has 1200 triangular faces, 720 edges and 120 vertices. The vertices of an origin-centered 600-cell with edges of length 1/p (where p = (1+sqrt(5))/2 is the golden ratio) can be given as follows:

2 cells are called adjacent, if they have at least 1 common vertex and 2 cells are called independent, if they have no common vertices.

Question: How many different ways is it possible to cover the vertices of the hypericosahedron by independent cells?

The topological structure of the 600-cell is a system of subsets over the 120 vertices that gives which k vertices form a k-facet (k = 2, 3, 4; 2-facets are called edges, 3-facets are called faces and 4-facets are called cells). The topological structure of the 600-cell can be computed easily: At first we generate the coordinates of the 120 vertices according to the previous scheme. Then we identify the k-facets by examining every possible k vertices and checking whether they are at distance 1/p from each other not. The result of this computation can be seen here: vertices, edges, faces, cells.

The cell adjacency graph of the 600-cell can be easily obtained from its topological structure. Now the original question can be converted to the following equivalent one: How many different ways is it possible to choose 30 independent points from the cell adjecency graph of the 600-cell? I wrote a simple program that answers this question using brute force computation. It could find 1920 different coverings.