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I want to confirm my solution to the given problem (solutions were not available in the book)

I have eight cubes. Two of them are painted red, two white, two blue and two yellow, but otherwise they are indistinguishable. I wish to assemble them into one large cube with each color appearing on each face. In how many different ways can I assemble the cube?

  1. The answer i got was 96 is this correct?
  2. I also tried to generalize the question such that given $n$ cubes and $\sqrt n$ number of colors, I came up with

$$n!\cdot2(n-\sqrt{n})!\cdot(n-2\sqrt{n})!\cdot2(n-(4\sqrt{n}-4))!$$

is the above generalization correct?

Thanks

EDIT

well to further explain my question, the rational behind the answer to the 1st question was we have 6 faces and if we take one of the faces we have 4! possibilities to arrange the 4 colors. that gives us 24 possibilities. and if move to the other faces, we have 2 faces of 2! possibilities and 3 faces of 1! possibilities thus total arrangements are 4!x2!x2!x1!x1!x1! = 96

for the second question yes the colors should be n^1/3 but for the generalization n should be no of colors.

for example if we take the current question

n=4

  1. if we take the front face as the base face we have n! = 24
  2. if we take the left face because of step 1 we have $(n-\sqrt n)! =2$
  3. if we take the back face because of step 2 we again have $(n-\sqrt n)! =2$
  4. if we take the right face because of step 1 and step 3 we have $(n-2\sqrt n)! =1$
  5. if we take the top face because of step 1,2,3,4 we have $(n-(4\sqrt n -4))! =1$
  6. if we take the bottom face again because of step 1,2,3,4 we have $(n-(4\sqrt n -4))! =1$

thus the result 4!x2!x2!x1!x1!x1! = 96

user119020
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  • Something feels off with the generalization: $4 \neq \sqrt{8}$, so I'm not sure what $\sqrt{n}$ is doing. I also don't know if this can be generalized by further subdividing a three-dimensional cube -- it seems more natural to make up a $d$-dimensional using $2^d$ cubes. But I'm not sure how many colors would be "right," or whether this generalization could work out either. But I like the thought to generalize! – pjs36 Apr 11 '16 at 05:36
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    The answer to (1) is "it depends on what counts as the same". I think there is an argument for the answer being two. Your generalisation in (2) gives an expression I cannot read, but anyway $9$ cubes and $3$ colours does not work – Henry Apr 11 '16 at 05:37
  • $4 \ne \sqrt{8}$ but $4 = 2^2; 8= 2^3$. To generalize your want $n^3$ cubes with $n$ cubes of $n^2$ number of colors. – fleablood Apr 11 '16 at 05:49
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    As in @browngreen's solution, once the colors on the front face are fixed, the arrangement of colors on the other five faces are uniquely determined. So you need to omit the two factors of $2!$. One way to think about it: if the colors on the front face are 1234, read in clockwise order from the top left, then, since the top face already uses 1 and 2, we need to add 3 and 4; since the right side face already uses 2 and 3, we need to add 1 and 4. But the top face and the right side face share an edge, so the missing color on that edge can only be 4. – Will Orrick Apr 11 '16 at 07:20
  • Another way to say it: one of the two possible arrangements of the left side face in your analysis is not going to work when you later go to color the top and bottom faces, but the other one will. – Will Orrick Apr 11 '16 at 07:27
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    I'm not sure the $\sqrt{n}\times\sqrt{n}\times\sqrt{n}$ generalization has been completely thought through. Just one possible issue: when $\sqrt{n}>2$, there will be interior cubes that are completely hidden from view, which may therefore be permuted arbitrarily. There are other problems as well. – Will Orrick Apr 11 '16 at 07:28
  • There are 4! ways to do the first face (without regard to orientation) but after that you have no choice at all. If you do the front face and then move move to any of the back pieces you have no choice, the back left top must be a different color than the front left top, front left bottom, or front right top, so it must be the same color as the front right bottom. In general (x,y,z) must be a different color than (x,#,#) and (#,y,#) and (#,#,z) so the only option for it is to be the same as (-x-y,-z). Every cube must be the same color as it's opposite corner. – fleablood Apr 11 '16 at 16:43
  • @WillOrrick I agree. That's why I suggest making a $d$-cube from $2^d$ colored cubes. Not sure how many colors we'd want, if it should be $2^{d - 1}$, or what. There's also the possibility of staying in $3$ dimensions; I guess we'd use $8 + 12(n - 2) + 6(n - 2)^2$ cubies, but I'm still not sure about the number of colors. – pjs36 Apr 12 '16 at 00:57
  • @ fleablood forgive my ignorance but im not sure why one has no choice after you have done the 1st face . let say we go clock wise and label the front face f1, f2,f3,f4 and the back face as b1,b2,b3,b4. now lets say we arrange the front face as f1:red, f2:green,f3:blue,f4:white. now the back face can be b1:green b2:red,b3:white b4:blue or b1:blue b2: white, b3: red, b4:green etc.. – user119020 Apr 12 '16 at 02:29
  • In the arrangement f1:red, f2:green,f3:blue,f4:white. b1:green b2:red,b3:white b4:blue, the top of the cube, consisting of f1, f2, b1, b2 has two reds and two greens. The bottom has two whites and two blues. – Will Orrick Apr 12 '16 at 03:23
  • @pjs36: I've posted an answer that contains one possible generalization to the $3\times 3\times 3$ cube. Indeed, there is one cube that is hidden from view. Nevertheless, one can find solutions in which all nine colors appear on every face. Higher $n$ may also be possible. – Will Orrick Apr 12 '16 at 16:43

7 Answers7

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If you are allowed to rotate the assembled cube freely there is exactly one way to assemble it.

Proof. Let $\{0,1,2,3\}$ be the set of colors. Color each vertex of the assembled $2$-cube with the color of the $1$-cube it belongs to. Vertices of the assembled cube having the same color are necessarily space-diagonally opposite. Assume that a vertex having color $0$ is facing you. Then its three adjacent vertices have colors $1$, $2$, $3$, arranged either counterclockwise or clockwise. Now convince yourself of the following: If $1$, $2$, $3$ are arranged counterclockwise, then $1$, $2$, $3$ will be arranged clockwise if you look at the assembled cube with the opposite $0$-vertex facing you.$\qquad\square$

It is so far unclear how this problem can be meaningfully formulated for an $n\times n\times n$ assembled cube.

Christian Blatter
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  • If I understand correctly, you approach this to color each face of the cube with a distinct color while the question is about color each face of the cube using all the available colors. I am assuming that by vertex you mean "face" that the question in the post is referring to. – Jim Jan 23 '21 at 17:28
  • @Jim: My approach is to assemble a $2\times2\times2$ cube as proposed by the OP. In addition spacial movements of the large cube are used to identify "analog" arrangements. – Christian Blatter Jan 23 '21 at 18:56
  • Yes but this cube has as base 4 1x1 cubes each cube of different color (all faces of 1 cube of 1x1 having one color), is that right? Then we would add the other 4 1x1 again of different color possibly rotated to end up with ${red, white, blue, yellow}$ on each face of the square. Turning it to the side we just see the sides of the original 1x1 cube. In that case wouldn't the arrangements be $4321$ for the base times $21$ for the top i.e. $432121 = 48$? It would be $2 1$ for the top or side 2 cubes of 1x1 since we will need to pick a different color than the adjacent one.Makes sense? – Jim Jan 23 '21 at 23:19
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This answer looks at Question 2, specifically the generalization to $n=3$.

Consider the problem of assembling a $3\times 3\times 3$ cube using a collection of $3^3=27$ cubies of $3^2=9$ colors, with $3$ cubies of each color, so that each color appears once on each face. There are four roles a cubie may play:

  • a corner cubie contributes its color to three faces;
  • an edge cubie contributes its color to two faces;
  • a center-of-face cubie contributes its color to one face;
  • the center-of-body cubie contributes its color to no faces.

Now each of the $9$ colors must appear on six faces. There are three ways to obtain $6$ as a sum of three numbers taken from $\{3,2,1,0\}$. They are $3+3+0$, $3+2+1$, and $2+2+2$. One of the nine colors must be used for the center-of-body cubie, so the other two cubies of that color must be diagonally opposite corner cubies (in order to have that color on all six faces). This leaves six other corners; since $3+3+0$ can only be used once, these remaining corners must each appear as part of the $3+2+1$ pattern, which means that each must be of a different color. This implies that the six center-of-face cubies must be of the same six different colors, as must six of the 12 edge cubies. There remain six edge cubies and two unused colors, which implies that the $2+2+2$ pattern must be used for the last two colors. That is, of the six remaining edge cubies, three are of one of the two remaining colors, and the other three are of the other remaining color.

Let the center-of-body cubie be of color $1$, and let the six center-of-face cubies be of colors $2$–$7$. We therefore start with the coloring $$ \begin{array}{ccccccccccccc} 1&x&x&\phantom{x}&\phantom{x}&x&4&x&\phantom{x}&\phantom{x}&x&x&x\\ x&2&x&&&6&1&7&&&x&3&x\\ x&x&x&&&x&5&x&&&x&x&1 \end{array}, $$ where the first $3\times3$ square represents the front face, the second $3\times3$ square represents the middle layer, and the third $3\times3$ square represents the back face, all viewed from the front. We see that, for example, the center of the front face has color $2$, which implies that one of the corner cubies on the back face must also have color $2$. There are three possible corners; once a corner is chosen, then the edge cubie with color $2$ is uniquely determined. We may proceed in this manner to find many ways of assembling the cube. A few of these are below.

Assembly 1: $$ \begin{array}{ccccccccccccc} 1&7&6&\phantom{x}&\phantom{x}&3&4&9&\phantom{x}&\phantom{x}&5&8&2\\ 8&2&5&&&6&1&7&&&9&3&4\\ 4&9&3&&&2&5&8&&&7&6&1 \end{array} $$

Assembly 2: $$ \begin{array}{ccccccccccccc} 1&7&6&\phantom{x}&\phantom{x}&9&4&3&\phantom{x}&\phantom{x}&5&8&2\\ 8&2&5&&&6&1&7&&&4&3&9\\ 3&9&4&&&2&5&8&&&7&6&1 \end{array} $$

Assembly 3: $$ \begin{array}{ccccccccccccc} 1&7&6&\phantom{x}&\phantom{x}&3&4&9&\phantom{x}&\phantom{x}&2&8&5\\ 5&2&8&&&6&1&7&&&9&3&4\\ 4&9&3&&&8&5&2&&&7&6&1 \end{array} $$

These three are certainly not equivalent, even up to permutation of colors and rotation. To see this, observe that in the third assembly the corner cubie of color $j=2,3,\ldots,7$ on the face opposite the face on which color $j$ appears as the center-of-face cubie is always diagonally across from color $1$, but that this is true of only four of the faces in the first assembly, and of only two in the second assembly.

Added: Up to rotation, reflection, and permutations of colors, the three assemblies above are the only ones possible. The fact is proved and then used to give a complete enumeration below.

As above, we initially suppose that the center-of-body cubie is of color $1$, which means that two body-diagonally opposite corners, which we will take to be front, top, left and back, bottom, right, are also of color $1$. Denote these corners $a$ and $b$.

Two colors, which we will suppose to be colors $8$ and $9$, are used only as edge cubies. Let $E$ be the set of three edges on which color $8$ appears. The $12$ edges of the cube come in three parallel classes of four edges each. No face of the cube may contain two edges of $E$ since each edge of $E$ must contribute two new faces to the total of six faces on which color $8$ is to appear. Furthermore, $E$ cannot contain two edges that are parallel and across the body center from each other, since the two faces that are left without color $8$ are then opposite each other, and have no edge in common that would allow color $8$ to appear on both. Hence $E$ must consist of one edge from each parallel class.

The edges of $E$ will be incident on a total of six corners and nonincident on two, the latter of which are necessarily body-diagonally opposite. (If the two non-incident corners were opposite on a face, then that face would not contain color $8$; if they were joined by an edge, then the two edges parallel to that edge and opposite to it on the coincident faces would have to contain color $8$, an arrangement that we have already ruled out.) The set $F$ of edges on which color $9$ appears has the same properties. The sets $E$ and $F$ are, of course, disjoint.

There are now three possibilities for the incidence of $a$ and $b$ on $E$ and $F$:

  1. neither $a$ nor $b$ is incident on any edge of $E$ or $F$;
  2. each of $a$ and $b$ is incident on an edge of one set, say $E$, and nonincident on any edge of the other;
  3. each of $a$ and $b$ is incident on an edge of both $E$ and $F$.

In case 1, since the three edges incident on $a$ and the three edges incident on $b$ are not used in $E$ or $F$, the set $E\cup F$ consists of all six remaining edges, which form a $6$-cycle. The edge of this $6$-cycle will alternate between the colors $8$ and $9$, giving two possible edge $2$-colorings, one of which is shown.

Case 1, with 6-cycle highlighted

After arbitrarily choosing one of the two possible edge $2$-colorings, say the one shown, and then arbitrarily assigning colors $2$–$7$ to the center-of face cubies, we have $$ \begin{array}{ccccccccccccc} 1&x&x&\phantom{x}&\phantom{x}&x&4&8&\phantom{x}&\phantom{x}&x&9&x\\ x&2&9&&&6&1&7&&&8&3&x\\ x&8&x&&&9&5&x&&&x&x&1 \end{array}. $$ We then find that the colors of all remaining cubies are forced, and the result is the third assembly above. As an example of this forcing: one of the three uncolored corners on the back face must receive color $2$. Whichever choice is made, we need to place one more cubie of color $2$, and it must be on an edge. Furthermore, the choice of edge is forced by the condition that every face contain the color $2$. We find that the color $2$ corner on the back face must be in the top left, since the other two choices would require the color $2$ edge cubie to be placed on an edge that is already occupied. Similar considerations force the placement of the remaining cubies of colors $3$, $4$, $5$, $6$, and $7$.

To enumerate the number of arrangements in this case, observe that we may color one face arbitrarily in $9!$ ways. We now need to decide which of the four body diagonals is to be the fixed-color one. Once this choice has been made, the rest of the arrangement is uniquely determined, giving $4\cdot9!$ arrangements.

In case 2, we choose one of the three edges incident on $a$ to be an element of $E$. The edge incident on $b$ parallel to this one cannot be in $E$, but one of the remaining two edges incident on $b$ must be chosen to be in $E$. The third element of $E$ is uniquely determined by these choices. We also find that the condition that no edge of $F$ be incident on $a$ or $b$ uniquely determines $F$ once $E$ has been specified. The edges of $E\cup F$ do not form a 6-cycle in this case. Instead, they form two disjoint paths of length $3$, one of which connects $a$ to $b$. An example is shown below.

Case 2

In this example we chose to put color $8$ on the front-left edge, incident on vertex $a$, and then on the right-bottom edge, incident on vertex $b$. The color $8$ was then forced on the back-top edge. Since color $9$ is not to be placed on an edge incident on $a$ or $b$, the $9$ on the front-bottom edge is forced. The other two $9$s are then also forced. Arbitrarily assigning colors $2$–$7$ to the six center-of-face cubies, we have $$ \begin{array}{ccccccccccccc} 1&x&x&\phantom{x}&\phantom{x}&x&4&9&\phantom{x}&\phantom{x}&x&8&x\\ 8&2&x&&&6&1&7&&&9&3&x\\ x&9&x&&&x&5&8&&&x&x&1 \end{array}. $$ The placement of colors $2$ and $3$ is then forced: $$ \begin{array}{ccccccccccccc} 1&x&x&\phantom{x}&\phantom{x}&3&4&9&\phantom{x}&\phantom{x}&x&8&2\\ 8&2&x&&&6&1&7&&&9&3&x\\ x&9&3&&&2&5&8&&&x&x&1 \end{array}. $$ Color $6$ needs to be used on a corner of the right face, only one of which is not yet assigned. This then forces color $6$ on the back-bottom edge. Similarly, the placement of color $5$ is forced: $$ \begin{array}{ccccccccccccc} 1&x&6&\phantom{x}&\phantom{x}&3&4&9&\phantom{x}&\phantom{x}&5&8&2\\ 8&2&5&&&6&1&7&&&9&3&x\\ x&9&3&&&2&5&8&&&x&6&1 \end{array}. $$ We now find that colors $4$ and $7$ can be placed in only one way, giving the first assembly above.

To enumerate the number of arrangements in case 2, observe that we may again color one face arbitrarily in $9!$ ways, and that we again need to decide which of the four body diagonals is to be the fixed-color one. Finally we need to choose among the six possibilities for the set $E$. Therefore in case 2 there are $4\cdot6\cdot9!$ ways of assembling the cube.

In case 3, we must choose the set $E$ as in case 2, but there are now three ways to choose the set $F$. (There are two ways to choose the edge in $F$ incident on $a$. If the chosen edge does not form a path with two edges of $E$, then the other two elements of $F$ are forced, as in the first picture below. If the chosen edge does connect two edges of $E$, then there are two ways to complete $F$, shown in the second and third pictures below.)

enter image description hereenter image description hereenter image description here

The first two of these are equivalent up to rotation of the cube; in these configurations, $E\cup F$ consists of two disjoint paths of length $3$. We find, however, that it is not possible to complete these configurations to a valid arrangement. For, taking the first one, we have $$ \begin{array}{ccccccccccccc} 1&9&x&\phantom{x}&\phantom{x}&x&4&x&\phantom{x}&\phantom{x}&x&8&x\\ 8&2&x&&&6&1&7&&&x&3&9\\ x&x&x&&&9&5&8&&&x&x&1 \end{array}. $$ We must now put color $2$ bottom-left corner of the back face and color $3$ on the bottom-right corner of the front face. Then color $4$ is forced to go on the front-left corner of the bottom face, which is impossible since then color $4$ would also have to go on the back-right edge, which is already occupied by color $9$.

In the third configuration, the edges of $E\cup F$ form a $6$-cycle containing corners $a$ and $b$, a condition that uniquely determines $F$ once $E$ is specified. Since there are six ways to choose $E$, there are six such 6-cycles, but they are equivalent in pairs under interchange of colors $8$ and $9$.

The configuration shown gives $$ \begin{array}{ccccccccccccc} 1&x&x&\phantom{x}&\phantom{x}&9&4&x&\phantom{x}&\phantom{x}&x&8&x\\ 8&2&x&&&6&1&7&&&x&3&9\\ x&9&x&&&x&5&8&&&x&x&1 \end{array}. $$ Color $6$, which must appear as a corner of the right face, can only go in the top-front corner. Likewise, color $7$ can only go in the back-bottom corner of the left face. Color $4$ must then go in the front-right corner of the bottom face and color $5$ in the back-left corner of the top face. The positions of colors $2$ and $3$ are then forced as well, giving the second assembly above.

To enumerate the number of arrangements in this subcase of case 3, once again we may color one face arbitrarily in $9!$ ways, and once again we need to decide which of the four body diagonals is to be the fixed-color one. Finally we need to choose among the three equivalence classes of $6$-cycles containing $a$ and $b$. Therefore in case 3 there are $4\cdot3\cdot9!$ ways of assembling the cube.

In conclusion, there are $4\cdot(1+6+3)\cdot9!$ arrangements of the $3\times3\times 3$ cube. For the $n\times n\times n$ cube, there seems to be a lot more flexibility when $n\ge4$, so I expect there to be many arrangements.

Will Orrick
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It depends on how you define different ways to assemble the cube. If there are two constructions of the cube such that I can get from one to the next just by rotating the cube without disassembling and assembling, is that considered two different cubes or one?

If you look at the large cube as 8 defined spots for the 8 small cubes, and a cube created by rotating is different from the cube that it was rotated from, then I would say there are 24 possible constructions. On any given side, there are 4 locations for say the white cube. Once I choose it's location, the other white cube must go on the opposite corner of the large cube. Then I have 3 locations for the next color and two for the next. So the total number of combinations is 24.

If however a cube created by rotating is still the same cube, then as far as I am aware, there is only one construction. By flipping on all 6 sides and turning in all 4 ways, I believe you can get all 24 possibilities mentioned earlier.

browngreen
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The requirement is to have each color on each face of the composed cube. What that means is that every pair of cubes of the same color must be arranged "diagonally" that is, touching corner to corner, or, in other words, if I put a red cube on the front-bottom-left then the other red cube must be placed on the rear-top-right. Is that clear until now? If not, then try to think why other arrangements won't satisfy the requirement above.

Now, apparently rotating the composed cube doesn't make it any different from the non-rotated version. With that in mind, let's see how many options we have to arrange the cubes on the "front" half. - At first it may seem that we have 4 options to choose where to place the first cube, and 3 options for the next one. And we can go that way... but there's an easier one:

Let's ask - how many ways are there to arrange 4 different colors on the "front" half. And the answer is 3*2. Why? Because it doesn't matter where we put the first chosen cube (whatever color it is) since we can rotate the cube. Therefore, what really matters, is how many ways are there to arrange the rest 3 colors. So we are left with 3 places for the second color, 2 places for the third, and only 1 for the last color. And what about the "back" half? It will arrange in a "flipped" and "mirrored" manner. And what about the sides - again - they are the mirrors of the "front". So we've found the simple method for counting only the "front" side and it greatly simplified our lives.

So the answer is 3! = 6.

valk
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Let's see. Label the eight positions FLB, FLT, FRB, FRT, bLB, bLT, bRB, bRT.

There 4 possible colors for FLB, 3 for FLT, 2 for FRB, 1 for FRT.

bLB cannot share its color with any L, of **B so the only color it can have is that of FRT. Each of the remaining three slots can only have the color of its opposite corner (Otherwise it will share a color with one of its three faces).

So there are 4! to color a 2x2 cube. Now there are 2 of each color so there are $2^4$ ways to chose which of the two cubes for each slot.

So there are $4!*2^4 = 24*16 = 384$.

But this distinguishes between rotations. If I distinguish that I will always start with, arbitrarily the first yellow cube in FLB position, and it must be adjacent to all three of the other colors and if I distinguish that, arbitrarily, a blue cube must be in FRB, then this is reduced by 8 (where the first yellow cube goes is 8 and what color to choose for FRB is 3) then there are 16 possibility.

If further the 2 cubes of the same color are indistinguishable then I must reduce by the choices of which blue, white and red cubes. Or $2^3$ leaving only 2 possibilities.

If we allow symmetries then there is only one possibility.

But if that were the interpretation there'd be simpler way to do the puzzle. The Yellow must go somewhere. On one of the faces the the yellow will be next to the blue (and on another face next to the yellow and the third next to the white). Orientate ourselves to that face and rotate it so the blue is to the right of the yellow. Our only choice is whether the red or the white should go above the yellow. This is essentially choosing whether we want a "right or left handed orientation".

To extend to an $n \times n \times n$ cube is different problem entirely as there are center unviewed cubes and not all cubes are on all three face.

To extend a 2 cube to the n-th dimension would be $2^{n-1}!*2^{2^{n-1}}$ and with respect to orientation $2^{n-1}!*2^{2^{n-1}}/2^n*n!$. (I think).

If we don't distinguish between individual cubes of the same color, I think, this will be $2^{n-2}$ which I believe are the number of left-handed/right-handed orientations. Or maybe it is only 2.

fleablood
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    The wording of the problem seems to suggest that the two red cubes are not distinguished from each other, and, in general, the two cubes of each color are not distinguished from each other. If one does choose to distinguish then, however, wouldn't fixing the identity of the cube in one of the corners reduce the count by $8$ rather than $4$? – Will Orrick Apr 11 '16 at 06:39
  • I'm also uncertain about the generalization. Are you assuming that diagonally opposite cubes are of the same color, so that in the $3\times3\times3$ cube, the center of the front face is the same color as the center of the back face, the middle cube of the top front edge is the same color as the middle cube of the bottom back edge, and so on, and top front left corner cube is the same color as the bottom back right corner cube, and so on? If so, then, after applying this idea to the coloring of the front and back faces, you run into conflicts when you try to color the middle layer of cubes. – Will Orrick Apr 11 '16 at 07:30
  • Duh. You are right. Obviously this doesn't extend beyond 2 . – fleablood Apr 11 '16 at 16:20
  • I thought the wording was entirely ambiguous, I thought. – fleablood Apr 11 '16 at 16:36
  • I was assuming diagonally opposite cubes are the same color or more that (x,y,z) is the same color as (x',y',z') where w and w' are reflections. I obviously wasn't thinking. That's true for n=2 because because w and w' are the only two options and (x',y'z') must be different from anything else with a x',y', or z' and anything else with a x',y', or z' must be have at least one x,y, or z and must be different from (x,y,z) so (x,y,z) and (x',y',z') must both be different to the same colors and must therefore be the same color. That obviously isn't the case if n $\ge$ 3. – fleablood Apr 11 '16 at 20:17
  • Is that a conflict with n = 2m? But that's another problem. – fleablood Apr 11 '16 at 20:30
  • In the $3\times3\times3$ case you can arbitrarily color the front face $$\begin{array}{ccc}1&2&3\ 4&5&6\ 7&8&9\end{array}.$$ With diagonally opposite cubes being the same color, the back face would be colored $$\begin{array}{ccc}9&8&7\ 6&5&4\ 3&2&1\end{array}.$$ Here the back face is being viewed from the front, as if the layers in front of it were transparent. Now both the left side face and the right side face already contain... – Will Orrick Apr 12 '16 at 00:29
  • ... colors 1, 4, 7, 9, 6, 3, which means that colors 2, 5, 8 need to be added (in the middle layer). But that means that two of the colors 2, 5, 8 will also touch the top face. This is a problem because colors 2 and 8 are already present on the top face. The same issue also occurs on the bottom face. One can, in fact, show without much difficulty that six different colors must be used on the center cubes of the six faces of the $3\times3\times3$ cube. I will post an answer with the proof. – Will Orrick Apr 12 '16 at 00:29
  • Well, any odd number obviously won't work the upper right hand corner of the middle layer will be the same as the lower left hand corner. – fleablood Apr 12 '16 at 00:45
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I will try to give a simple proof, by using the techniques taught in the book itself.

ENTRY Phase

  • Draw a diagram

enter image description here

  • Write down what you know

    • There are 8 cubes which are identical to each other, except for their colors. There are 2 red, 2 green, 2 blue and 2 orange cubes.

  • Be clear on what you want

    • We want to assemble as many larger 2X2 cubes from these 8 smaller cubes with the following constraint - each color should appear on each face.
    • Goal is to find the number of assemblies possible.

Specialize

Can you solve one particular case? Yes - as shown in the picture.

What else you notice? You could observe that cubes with same color are at diagonally opposite corners of the cube.

Also observe a hidden constraint that for any face to have each of the 4 colors, you need to have each of the color on all of the faces.

Attack

Can you conjecture based on the observation?

Specialize further by refining.

Conjecture:

To satisfy the given constraint, cubes of similar color has to be on diagonally opposite corners of the cube.

Visual Proof:

Let us represent positions of the smaller cubes using following matrix, with first matrix for the front face and second one for the rear face, both read from left to right. $$\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} \begin{bmatrix}5 & 6\\7 & 8\end{bmatrix}$$

  • If 2 cubes are next to each other on the same row in a face, that will violate the constraint on two faces. So (1,2) or (3,4) cannot be of same color. By symmetry, this is true for rear face as well. ie, (5,6) or (7,8) also can't be of same color.
  • If 2 cubes are diagonally opposite to each other on the same face, that will violate the constraint on that face. So (1,4) or (2,3) cannot be of same color. By symmetry, this is true for rear face as well. ie, (5,8) or (6,7) also can't be of same color.
  • By symmetry after rotation, it can be argued that 1,6 can't be of same color. Similarly, (2,5), (3,8) and (4,7) can't be of the same color.
  • Therefore same color pieces have to be in positions (1,8), (2,7), (3,6) and (4,5).

If we fix a given color, say blue at position 1, there are 24 combinations possible. If we swap blue with 3 other colors, there are another $24 \cdot 3$ possibilities.

So a total of $24 \cdot 4 = 96$ combinations are possible, which is the final answer.

senseiwu
  • 214
-1

Yes think like this to make a cube max side is $2$ as $2^3=8$ qnd we have $8$ eight cubes. Now take tge base . The four cubes can be arranged in $4!$ ways in the base. Now as we want each colour on each face $2$ cubes with colour different than base colour cubes can be arranged on top of two base cubes maybe row or column. So now they can be arranged in $2$ ways and same can be done with the cubes in other row or column .so these are $2$ ways. Hence total ways are $4!.2.2=96$ and may be number of colours in generalized formula should be $n^{1/3}$

Archis Welankar
  • 15,910