Let $P = \operatorname{conv}\lbrace (s(1), s(2), s(3), s(4)) \mid s \in S_4 \rbrace$ be a permutahedron. Compute the 3-dimensional measure of this polytope.
I know that $P$ is three dimensional (this is easy), but I don't have an idea how to proceed. Can you give me a hint please?
Hint: $P$ is cut from the octahedron with vertices at $(1,1,4,4)$, $(1,4,1,4)$,..., $(4,4,1,1)$.
Added from comments: One of the faces of $P$ is included in $x_1 = 4$. The vertices on that face are $(4,1,2,3)$, $(4,2,1,3)$, ... Those six vertices form a regular hexagon obtained by cutting the corners of the equilateral triangle with vertices at $(4,1,1,4)$, $(4,1,4,1)$, and $(4,4,1,1)$. The opposite face is included in $x_1 = 1$. Another kind of face is included in $x_1+x_2 = 7$. The vertices here are $(4,3,1,2)$, $(4,3,2,1)$, $(3,4,2,1)$, and $(3,4,1,2)$. That is a square and is the face resulting from the cut.
The correct picture of the embedded permutahedron is actually the following: https://en.wikipedia.org/wiki/Permutohedron#/media/File:Symmetric_group_4;_permutohedron_3D;_permutations_and_inversion_vectors.svg