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This question has likely been answered in full detail before, so any references would be greatly beneficial. The question I have is as follows:

Suppose we have $m$ central hyperplanes in $\mathbb{R}^n$, that is, $m$ hyperplanes of dimension $n-1$ that cross through the origin. How many regions does this arrangement split $\mathbb{R}^n$ into?

It isn't difficult to find the answer for noncentral arrangements, this is given by Zaslavsky's Theorem. Furthermore there is a detailed treatment of hyperplane arrangements in "An Introduction to Hyperplane Arrangements" at https://www.cis.upenn.edu/~cis610/sp06stanley.pdf. However, I wasn't able to find the result I am looking for.

What I could find is that if we allow the hyperplanes to be in general position, then the arrangement $A$ of $m$ hyperplanes in $\mathbb{R}^n$ has \begin{equation} r(A) = \sum_{j=0}^n {m \choose j} \end{equation} regions.

For the central arrangement question, the interesting case is when $m > n$. I believe that when $m \leq n$, the number of regions is simply $2^m$.

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When $n\ge 1$ and $m\ge 1$, the answer is $$2\sum_{j=0}^{n-1}\binom{m-1}j.$$To derive this from the non-central answer, given $m$ central hyperplanes in $\mathbb R^n$, let $P$ be one of the planes. Define two planes $P^+$ and $P^-$ which are parallel to $P$, such that the origin is between $P^+$ and $P^-$. Every region is on one side of $P$ or the other, so it either passes through $P^+$ or $P^-$. The plane $P^+$ is $(n-1)$ dimensional, so it is divided by the other $m-1$ planes into $\sum_{j=0}^{n-1}\binom{m-1}j$ regions, by the previous result. These regions of $P^+$ correspond exactly to the original regions which are on the $P^+$ side. You then multiply by $2$ to also account for the $P^-$ regions.

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  • $\begingroup$ Yes this is what I determined as well, thanks the explanation helps a ton too! $\endgroup$ – Merkh Jun 25 at 15:50

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