A real HARD problem of DIMENSIONS!

We first divide the set of simple hyperplans into four subsets, depending whether the hyperplan has 1, 2, 3 or 4 non-zero coefficients. We'll call an hyperplan with $n$ non-zero coefficients an hyperplan of order $n$ . We will also notice that the number of regions that the unit ball is divided into is the same as the number of regions that $\mathbb{R}^4$ is divided into, since multiplying $x_1$ , $x_2$ , $x_3$ and $x_4$ by some positive real number don't change the region into which this point fall.

There are 4 hyperplans of order 1: $x_1=0$ , $x_2=0$ , $x_3=0$ and $x_4=0$ . It is pretty simple to see that those 4 hyperplans divide the space into $2^4=$ 16 regions , since every hyperplan doubles the number of regions.

Because of symmetry, hyperplans of order 2 divide each of these 16 regions into the same number of regions. Hence, we can focus only on one region and multiply by 16. We'll focus on the region where $x_1>0$ , $x_2>0$ , $x_3>0$ and $x_4>0$ . There are 12 hyperplans of order 2, but only 6 of them pass through this region: $x_1-x_2=0$ , $x_1-x_3=0$ , $x_1-x_4=0$ , $x_2-x_3=0$ , $x_2-x_4=0$ and $x_3-x_4=0$ . Each region delimited by those hyperplans correspond to a permutation of the 4 variables, so there are $4!=$ 24 regions .

For order 3, it's a little bit more difficult. We'll focus on the region $0<x_1<x_2<x_3<x_4$ ; out of the 16 hyperplans of order 3, only 4 pass into this region: $x_1+x_2-x_3=0$ , $x_1+x_2-x_4=0$ , $x_1+x_3-x_4=0$ and $x_2+x_3-x_4=0$ . We have to find the number of ways to assign $>0$ or $<0$ to each of the four equations, while respecting the following constraints, deduced from the assumption that $0<x_1<x_2<x_3<x_4$ :

$(x_1+x_2-x_4)<(x_1+x_3-x_4)<(x_2+x_3-x_4)$

$(x_1+x_2-x_4)<(x_1+x_2-x_3)$

If $x_1+x_2-x_4$ is positive, all terms are forced to be positive. If it is negative, the other terms are free to be positive or negative, except that if $x_1+x_3-x_4$ is positive, $x_2+x_3-x_4$ must also be positive. Hence, we count 7 ways to assign $>0$ or $<0$ to the four terms, so the region on which we were focusing is divided into 7 regions .

Finally, for hyperplans of order 4, we will focus on the region determined by $0<x_1<x_2<x_3<x_4$ and $(x_1+x_2-x_3), (x_1+x_2-x_4), (x_1+x_3-x_4), (x_2+x_3-x_4)<0$ . There are 8 hyperplans of order 4, but we can verify that only one of them pass through this region: $x_1+x_2+x_3-x_4=0$ . Hence, the region is divided into 2 regions .

We can now conclude that the set of all simple hyperplans divide $\mathbb{R}^4$ into $16*24*7*2=$ $\boxed{5376}$ regions.

N.B. It would also be interesting to find a general formula for simple hyperplans in $\mathbb{R}^n$ . I found the first values: 2 for $n=1$ , 8 for $n=2$ , 96 for $n=3$ and 5376 for $n=4$ , but I'm not able to find a simple formula for higher dimensions.