This non-answer completes Joseph O'Rourke's nice non-answer, for the case  of $n$ **hyperplanes** in $\mathbb{R}^d$ in general position. But it also suggests that the OP situation may also well have unique answers.

Define:

 $U_{d,n}=$ number of unbounded regions cut by $n$ hyperplanes in $\mathbb{R}^d$

 $B_{d,n}=$ number of bounded regions

 $T_{d,n}=$ total number of regions $=U_{d,n}+B_{d,n}$

$S_{d,n}=$ number of regions cut on the sphere $S^d$ by $n$ great $S^{d-1}$-circles

Then $U_{d,n}, B_{d,n}$, $T_{d,n}$ and $S_{d,n}$ are unique with these formulas:

$U_{d,0}=1$, $\quad B_{d,0}=0$, $\quad T_{d,0}=1$, $\quad S_{d,0}=1$

$U_{1,n}=2$, $\quad B_{1,n}=n-1$, $\quad T_{1,n}=n+1$

and for $n>0$

1. $U_{d+1,n}=S_{d,n}$
2. $S_{d,n}=U_{d,n}+2B_{d,n}$
3. $T_{d,n+1}=T_{d,n}+\sum_{i=0}^{i=d-1}{n\choose i}$

*Proof*.

1. In $\mathbb{R}^{d+1}$ take a huge and growing $S^d$ sphere, so that all the bounded regions zoom down to a point at the center of the sphere, the hyperplanes become great circles on the sphere and the unbounded regions corresponds to regions cut by the circles on the sphere. Therefore *if the numbers are unique* (as will be proved at the end) 1. follows.

2. Centrally project $\mathbb{R}^d$ onto a half $S^d$ (tangent to it). Complete the semisphere to a sphere by central symmetry. Then the hyperplanes become great circles, the $B_{d,n}$ bounded regions in $\mathbb{R}^d$ become $2B_{d,n}$ regions in $S^d$ and the $U_{d,n}$ unbounded ones become $U_{d,n}$ regions stretching across the suture line (equator) of the sphere. Again by unicity 2. follows.

3. Start with $\mathbb{R}^d$ and $n$ hyperplanes in generic position inside it. Now add a new hyperplane in generic position the following way:
first chose a point inside one region: no matter how that point is eventually stretched to a hyperplane, to it will split the region in two, for a gain of 1, or $n \choose 0$. Now stretch that point to a line: since it is a generic line it will meet each of the $n$ hyperplanes once and at each meeting the line will cross into one one new region and split it - with a gain of $n \choose 1$ new regions. Next stretch the line to a generic 2-plane, which will meet once each of the $(d-2)$-dimensional intersections of two hyperplanes; at each meeting the growing plane will arrive from having already crossed 3 of the 4 regions, to cross into the fourth and cut it; this a gain of another $n \choose 2$ regions.
In general as a generic $m-1$-plane grows to a generic $m$-plane it will meet all the $n \choose m$ $(n-m)$-dimensional intersections of $m$ hyperplanes, and each time it will go from cutting $2^m-1$ regions before crossing the intersection to cutting all $2^m$ after crossing, for a total gain of $n \choose m$ regions. This continues up to $m=d-1$, proving 3.

***Proof of Unicity***. 

By induction:

$U_{d,0}$, $\quad B_{d,0}$, $\quad T_{d,0}$, $\quad S_{d,0}$ are **unique**;

$T_{d,n}$ **unique** $\implies$ $T_{d,n+1}$ **unique** (by the proof of 3.);

$U_{d,n}$ and $B_{d,n}$ **unique** $\implies$ $S_{d,n}$ **unique** (by the proof of 2. and the fact that the construction can be reversed in a non-unique way to show that $S_{d,n}=U_{d,n}+2B_{d,n}$ for *some* values of $U_{d,n}$ and $B_{d,n}$);

$S_{d,n}$ **unique** $\implies$ $U_{d+1,n}$ **unique** (by the proof of 1.);

$U_{d+1,n}$ and $T_{d+1,n}$ **unique** $\implies$ $B_{d+1,n}$ **unique** (as $B=T-U$).