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Seeking Patterns

Advanced students can express these results in a general formula. Let ©(k,n) denote the number of k-cubes in an n-cube. To calculate ©(k,n) we begin, as before, by counting how many k-cubes there are at each vertex. Each k-cube is determined by a subset of k distinct edges from among the n edges emanating from each vertex. Therefore the number of k-cubes at each vertex is C(k,n) = (n choose k) = n!/(k! (n-k)!), the combination of n things taken k at a time. Since there are C(k,n) k-cubes at each of the 2n vertices, the total number of k-cubes appears to be 2nC(k,n). But in this count each k-cube is counted 2k times, so we divide by that number to get the final formula: ©(k,n) = 2n-k C(k,n).

Remembering the pattern of powers of 2 that come from the sums of rows in the simplex table, we naturally seek a similar pattern for cubes. In this case the entries in each row add up to a power of 3:

DIMENSION: 0-cubes
(points)
Vertices
1-cubes
(lines)
Edges
2-cubes
(squares)
Faces
3-cubes
(cubes)
Cubes
4-cubes
(hypercubes)
4-Cubes
Sum

Point: 1 0 000 1
Line: 2 1 000 3
Square: 4 4 100 9
Cube: 812 61027
Hypercube:1632248181

There are several ways to react to this observation. We can generate an additional row of the table to gain some additional information, but the conjecture is fairly firmly established with the five completed rows. We can observe that each entry is the sum of twice the entry directly above it plus the entry to the left of that one, so the sum of entries in one row is three times the sum of entries in the previous row — an argument that can easily be translated into a formal proof by mathematical induction. We may also use the explicit formula for the number of k-cubes in an n-cube, to sum a typical row:

[HELP]

©(0,n) + ©(1,n) + للل + ©(n-1,n) + ©(n,n)
=2n + C(1,n)2n-1 + C(2,n)2n-2 + للل + C(n-1,n)2 + C(n,n)
=(2 + 1)n = 3n

All these approaches help explain why the rows sum to power of 3. But perhaps the most satisfying observation that justifies this fact is that we may divide the sides of an n-cube into three equal parts whose projections divide the entire cube into 3n small cubes (Figure 46). The result is a small cube coming from each vertex of the original cube, one from each edge, one from each two-dimensional face, and so on. The final small cube is in the center. Thus the total number of small n-cubes, which is 3n, is equal to the sum of the number of k-cubes in the n-cube — since there is one small n-cube for each point, edge, face, 3-cube, etc.

Figure 46. Subdivision of the sides of segments, squares, and cubes (and even hypercubes) into three equal parts yields 3, 9, 27, or 81 similar small objects — always powers of 3.

One of Friedrich Froebel's kindergarten gifts was a cube subdivided into 27 small cubes. He would have liked this final demonstration. [an error occurred while processing this directive]