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I know that in dimension $n \geq 5$ there are only 3 kind of convex regular polytopes in each dimension: the $n$-simplex, the $n$-cube and the $n$-orthoplex.

I would like to know if there are formulas that give the number of vertices, edges, faces, cells, 4-faces, 5-faces, 6-faces, 7-faces, ..., $n$-faces, of each of these convex regular polytopes.

Also, I heard that convex regular polytopes in dimension $n \geq 5$ only have triangles or squares as their faces. Is that true?

And what about their cells? Are they also all necessarily tetrahedrons (3-simplex), cubes (3-cube) and octahedrons (3-orthoplex)?

And what about their $n$-faces?

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The $n$-simplex has $\binom{n+1}{k+1}$ faces of dimension $k$.

Then $n$-cube has $2^{n-k}\binom{n}{k}$ faces of dimension $k$.

The $n$-crosspolytope (aka $n$-orthoplex) is dual to the $n$-cube, so it has $2^{k+1}\binom{n}{k+1}$ faces of dimension $k$ (the number of $(n-k-1)$-faces of the cube.)

Yes, all of these polytopes have only triangles or squares as 2-faces; in fact, all the faces of the cube are lower-dimensional cubes, and all the faces of the simplex or the crosspolytope are lower-dimensional simplices. (In fact, every convex polytope of dimension at least five has at least one 2-face which is a triangle or a quadrilateral.)

So the cells of hypercubes are 3-cubes, and cells of simplices or crosspolytopes are tetrahedra.

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    $\begingroup$ ... which are proved by induction, the inductive steps being simplexcount(n,k) = simplexcount(n-1,k)+simplexcount(n,k-1) and cubecount(n,k) = 2*cubecount(n-1,k)+cubecount(n-1,k-1) respectively. $\endgroup$ – Don Hatch Jul 13 '18 at 11:24

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