Consider a diamond, with the four corners at (1,0), (-1,0), (0,1), and (0,-1). The three-dimensional equivalent of this diamond would be an octahedron with the additional vertices (0,0,1) and (0,0,-1). The four-dimensional equivalent would have two more vertices (0,0,0,1) and (0,0,0,-1). In general, call the resulting shape an n-hyper-diamond.

If you have an n-hyper-diamond, how many m-dimensional hyper-faces does it have (where n>m≥0)? For example, in the case where n=3, an octahedron, there are 8 faces (m=2), 12 edges (m=1), and 6 vertices (m=0).