How many distinct patterns are possible if you omit (a) 1 piece, (b) 2 pieces and (c) 3 pieces from a cube originally consisting of 27 smaller and equally sized cubes?
[Math] Cube stack problem
geometry
Related Solutions
Here's my hint: Stare at that for a sufficiently long time, and the solution will come up.
and if that doesn't help, try this:
For the third method you need to do faces (1 side painted) - edges (2 sides painted) + vertices (3 sides painted). The idea is that the edges get counted twice by each face so we fix by subtracting that total back out. But then the vertices (corners of the cube) end up not being counted at all since each vertex is hit by 3 faces and 3 edges have we have just done 3 - 3. So we have to add back in the number of vertices.
There are 6 faces with 10*10 cubes which gives 600. There are 12 edges with 10 cubes which gives 120. There are 8 vertices (corners) with a single cube each which gives 8. So we have $600-120+8 = 488$.
Best Answer
We will solve this problem with the Polya Enumeration Theorem. Evidently this is equivalent to counting two-colorings of the small cubes (leaving out a cube is just like coloring it a second, different color). This requires the cycle index of the permutation group $G$ of the small cubes inside the large one. It is important to note that we restrict ourselves to rotations. If we had a graph instead of a real-world object there would be mirror reflections to consider as well (reflections about a plane passing through the center of the cube and parallel to two opposite sides). This cannot be done with a physical concrete object. We now enumerate the permutations of $Z(G)$ by their cycle structure, considering the types of rotations in turn.
Start with the identity, which contributes $$a_1^{27}.$$
Now there are eight rotations about axes passing through opposite corners of the cube (two rotations per axis, of which there are four), which fix the three cubes on the diagonal and which contribute $$8\times a_1^3 a_3^8.$$ There are three rotations about axes passing through the centers of opposite faces, which again fix the cubes on said axis and which give $$3\times (2 a_1^3 a_4^6 + a_1^3 a_2^{12}).$$ Finally we have six rotations about axes passing through the centers of opposite edges, which fix the cubes on that axis and give $$6\times a_1^3 a_2^{12}.$$ This gives the cycle index $$Z(G) = \frac{1}{24} \left(a_1^{27} + 8\times a_1^3 a_3^8 + 3\times (2 a_1^3 a_4^6 + a_1^3 a_2^{12}) + 6\times a_1^3 a_2^{12}\right)$$ which is $$Z(G) = \frac{1}{24} \left(a_1^{27} + 8\times a_1^3 a_3^8 + 6\times a_1^3 a_4^6 + 9\times a_1^3 a_2^{12}\right).$$ Now this gives $$Z(G)(1+z) = 1/24\, \left( 1+z \right) ^{27}+1/3\, \left( 1+z \right) ^{3} \left( 1+{z}^{3} \right) ^{8}\\+1/4\, \left( 1+z \right) ^{3} \left( 1+{z}^{4} \right) ^{6}+3/8\, \left( 1+z \right) ^{3} \left( 1+{z}^{2} \right) ^{12}$$ which is $$Z(G)(1+z) = {z}^{27}+4\,{z}^{26}+22\,{z}^{25}+139\,{z}^{24}+779\,{z}^{23}+3455\,{z}^{22}\\+12507\, {z}^{21}+37303\,{z}^{20}+92968\,{z}^{19}+195963\,{z}^{18}+352433\,{z}^{17}+544382\,{ z}^{16}\\+725612\,{z}^{15}+837184\,{z}^{14}+837184\,{z}^{13}+725612\,{z}^{12}+544382\, {z}^{11}+352433\,{z}^{10}\\+195963\,{z}^{9}+92968\,{z}^{8}+37303\,{z}^{7}+12507\,{z}^{ 6}+3455\,{z}^{5}\\+779\,{z}^{4}+139\,{z}^{3}+22\,{z}^{2}+4\,z+1,$$ so the answers are four, twenty-two, and one hundred thirty nine respectively.
This MSE link computes the full face permutation group of the n-dimensional hypercube.