Triangulation of hypercubes into simplices 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.)

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.
