Puzzle About Cubes (from the book thinking mathematically) 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?



*

*The answer i got was 96 is this correct? 

*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


*

*if we take the front face as the base face we have n! = 24

*if we take the left face because of step 1 we have $(n-\sqrt n)! =2$

*if we take the back face because of step 2 we again have $(n-\sqrt
    n)! =2$

*if we take the right face because of step 1 and step 3 we have
$(n-2\sqrt n)! =1$

*if we take the top face because of step 1,2,3,4  we have $(n-(4\sqrt
    n -4))! =1$

*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
 A: 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.
A: 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.
A: 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.
A: 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.  
A: I will try to give a simple proof, by using the techniques taught in the book itself.
ENTRY Phase

*

*Draw a diagram



*

*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.
A: 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 $n^{1/3}$
