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8 units cubed, 24 units squared and 32 units
Consider a normal unit cube. Placed on a cartesian coordinate system, each vertex has coordinates with a combination of ones and zeroes. Edges consist of two points, and the two points' coordinates differ in one of the numbers but not in the other two numbers. This is because the edge only has one dimension. The edge can use either the x, y or z dimension, and the two unused dimensions each can have values of either 0 or 1. There are twelve edges because twelve is 3*22.
The faces of the cube are the same as edges, except they use two dimensions. They can use any two of the three dimensions, and the last dimension has either a 0 or 1. There are six faces because six is 3*2. The same can be done with a hypercube.
In a 4-d hypercube, edges can use any one of the four dimensions. Each of the last three dimensions has either a 0 or 1. The number of edges is then 4*23, or 32.
Each two dimensional face of the hypercube can use any combination of two the four dimensions. There are C(4,2), or 6 such combinations. Each of the last two dimensions has a 0 or 1. The total number of faces must be C(4,2)*22, or 24.
The number of 3-d hyperfaces is calculated the same way. There are only 4 combinations of 3 dimensions that the hyperfaces can use. There is only one unused dimension. The number of hyperfaces is 4*21, or 8.
The equation that I used and explained is C(c,s)*2c-s, where C(c,s) is the number of possible combinations of used dimensions, and the exponent c-s is the number of unused dimensions. |
