Plane dividing 4d space I have a 4d space divided into 16ths about the origin. e.g. $(1,1,1,1)$ is contained in one 16th, $(-1,-1,-1,-1)$ is contained another, $(1,1,-1,-1)$ in a third.  If I add a plane to my 4d space, what is the maximal number of 16ths that my plane can pass through?
For example, if I have a 3d space divided into 1/8s, I am able to get a plane to pass through 7 of the 8 regions.  I'm not sure if this relates to the third row of Pascal's Triangle: 
($\binom30+\binom31+\binom32 = 1+3+3 = 7$) or just a coincidence.  
 A: If you divide $n$-dimensional space using $n$ hyperplanes ($n-1$ -dimensional subspaces; see the Wikipedia article on hyperplanes for more information), each perpendicular to an axis, and all intersecting at origin, you'll get $2^n$ cells defined by the hyperplanes.
In essence, you can label each cell by considering negative coordinates $0$, positive $1$ (choose either one for zero), and writing the valid ranges of each Cartesian coordinate axis as a binary number. For example, in 3D:
$$\begin{array}{c|c|c}
 z & y & x & \text{Binary} & \text{Decimal} \\ \hline
\lt 0 & \lt 0 & \lt 0 & 000 & 0 \\
\lt 0 & \lt 0 & \le 0 & 001 & 1 \\
\lt 0 & \le 0 & \lt 0 & 010 & 2 \\
\lt 0 & \le 0 & \le 0 & 011 & 3 \\
\le 0 & \lt 0 & \lt 0 & 100 & 4 \\
\le 0 & \lt 0 & \le 0 & 101 & 5 \\
\le 0 & \le 0 & \lt 0 & 110 & 6 \\
\le 0 & \le 0 & \le 0 & 111 & 7 \\
\end{array}$$
If you have more dimensions, just add new axes, and new binary digits. $n$ binary digits can represent exactly $2^n$ distinct values.
A hyperplane with all its normal components $\pm 1$ and slightly above or below the origin will pass through all but one cell; through $n^2-1$ cells.
If you have one dimension, you have $2^1 = 2$ cells. Each hyperplane is a point, and can be in either cell.
If you have two dimensions, you have $2^2 = 4$ cells. Each hyperplane is a line, and you can draw a line through three of the cells (diagonally, just above or below the origin).
If you have three dimensions, you have $2^3 = 8$ cells. Each hyperplane is a plane, and you can place it to intersect seven of the cells.
I do not have a mathematical proof of this. (If anyone does, I'd like to see it as a separate answer.) I do believe that a proof is possible using the
$$c_0 + \sum_{i=1}^n c_i x_i = 0$$
form, and enumerating the rules in which the hyperplane can pass through each cell. In particular, if $c_0 = -1$ and $c_i = 1$, the hyperplane cannot pass through the 0 cell (with all coordinates negative), but it does pass through all the other cells.
