Triangulation of hypercubes into simplices

Question

A square can be divided into two triangles. A 3-dimensional cube can be divided into 6 tetrahedrons. Into what number of simplices an n-dimensional hypercube can be divided? (For example, a 4-hypercube, or a 5-hypercube.)

Answer 1

Noting that the volume of a cone-like object $$O = \bigcup_{t\in[0,h]} \{t\}\times \frac thA$$ Is given by $\mathrm{Vol}^n(O) = \frac hn \mathrm{Vol}^{n-1}(A)$ suggests that we need $n$ $(n-1)$-simplices to get from a $n-1$-hypercube (hyper-parallelepiped) to a $n$-hypercube, i.e. $$N(n) = n \cdot N(n-1); \quad N(1) = 1 \Rightarrow N(n) = n!$$ Note that this only constitutes an upper bound. For 3D it gives $6$ when there is in fact a triangulation with only $5$ simplices.

Answer 2

As mentioned before there are (unsurprisingly) many ways to triangulate the cube. However one easy (and sometimes useful) way to do this is: Let $\pi$ be a permutation of $\{1,2,...,d\}$. Then define $S_\pi = \{x \in \mathbb{R}^d: 0 \leq x_{\pi(1)} \leq x_{\pi(2)}\leq ... \leq x_{\pi(d)} \leq 1 \}$. Clearly, $S_\pi$ is a simplex since it is described by $d+1$ linearly independent inequalities. Now the cube $[0,1]^d$ can be triangulated by all $S_\pi$ where $\pi$ ranges over all permutations of $\{1,2,...,d\}$. To check that these simplices have disjoint interior is an easy exercise. This method uses $d!$ simplices corresponding to $2! = 2$ in dimension 2 and $3!=6$ in dimension 3.