How to decompose a unit cube into tetrahedra?
The simplest way is a partition into $3! = 6$ tetrahedra:


$\{x,y,z : 0 \leq x \leq y \leq z \leq 1\}$.
$\{x,y,z : 0 \leq x \leq z \leq y \leq 1\}$.
$\{x,y,z : 0 \leq y \leq x \leq z \leq 1\}$.
$\{x,y,z : 0 \leq y \leq z \leq x \leq 1\}$.
$\{x,y,z : 0 \leq z \leq x \leq y \leq 1\}$.
$\{x,y,z : 0 \leq z \leq y \leq x \leq 1\}$.


The corresponding tetrahedra have the following vertices:


$(0,0,0),(0,0,1),(0,1,1),(1,1,1)$.
$(0,0,0),(0,1,0),(0,1,1),(1,1,1)$.
$(0,0,0),(0,0,1),(1,0,1),(1,1,1)$.
$(0,0,0),(1,0,0),(1,0,1),(1,1,1)$.
$(0,0,0),(0,1,0),(1,1,0),(1,1,1)$.
$(0,0,0),(1,0,0),(1,1,0),(1,1,1)$.