Permutahedron of three vectors (1,1,0,0), (−1,1,0,0), (−1,−1,0,0). 
I'm getting stuck on parts b, c, and d. Since visualizing the polytope is not possible, I think the way to find the facets and edges of P is to determine which combinations of points form facets and edges, respectively. 
For part b, I don't know how to work with those 4-d vectors to form 3-d shapes. Likewise for part d, I can't think of a way to see which pair of vertices would form an edge.
For part c, I think that once I can determine the number of edges of P, I can use Euler's equation ($V - E + F = 2$) to find the number of facets of P. But again this requires understanding how edges of P are formed.
 A: The permutachoron (infact, a 4D polytope is being called a polychoron, while a 3D polytope is being called a polyhedron) of the points (+1,0,0,0) and (-1,0,0,0) quite obviously is the 4D cross-polytope, also known as hexadecachoron or 16-cell. Therefore the searched for permutachoron is nothing but the rectified hexadecachoron.
When providing the hexadecachoron in $D_4$ symmetry its Coxeter-Dynkin digram is x3o3o *b3o (I'm using here typewriter-friendly linearisation of the tri-dental graph, esp. means that "*b" part a virtual revisiting of the b-th real node so far being mentioned at its left). When representing that hexadecachoron in hypercubical symmetry $BC_4$, then it would be given as x3o3o4o. Accordingly, when asking for the rectified hexadecachoron, then we'd get the (alternative) symbols o3x3o *b3o resectively o3x3o4o.
It just happens that those 2 latter symbols (resp. its being encoded polychoron) also can be represented within $F_4$ symmetry as x3o4o3o. In fact it is the icositetrachoron or 24-cell. All its cells are being given variously as .3x3o *b3o and o3x3o *b3. and o3x3. *b3o (wrt. $D_4$), as .3x3o4o and o3x3o4. (wrt. $BC_4$), or as x3o4o3. (wrt. $F_4$), all of which are representations of octahedra. The vertex figure of the icositetrachoron clearly is variously x3.3x *b3x (wrt. $D_4$), or x3.3x4o (wrt. $BC_4$), or .3x4o3o (wrt. $F_4$), that is the cube.
The incidence matrix of the hexadecachoron or 4D cross-polytope clearly is given as
8 |  6 | 12 |  8
--+----+----+---
2 | 24 |  4 |  4
--+----+----+---
3 |  3 | 32 |  2
--+----+----+---
4 |  6 |  4 | 16

The according incidence matrix of the searched for icositetrachoron likewise is given by
24 |  8 | 12 |  6
---+----+----+---
 2 | 96 |  3 |  3
---+----+----+---
 3 |  3 | 96 |  2
---+----+----+---
 6 | 12 |  8 | 24

--- rk
