In the picture, there is a cube with an edge of $2 \mathrm{~cm}$ made up of eight smaller cubes with an edge of $1 \mathrm{~cm}$. Eight faces of the smaller cubes are painted black, and the rest are white. From them, a cube can be assembled whose surface is white. In how many ways can the smaller cubes be painted? Assume that identically painted smaller cubes cannot be distinguished and can therefore be interchanged.

![](https://cdn.mathpix.com/cropped/2024_04_17_2a0500418bf8c778174dg-4.jpg?height=177&width=203&top_left_y=551&top_left_x=961)

(K. Pazourek)
From the task, it follows that each of the eight small cubes, from which the large cube is composed, has three white faces that share a common vertex. The remaining faces of each small cube are either black or white. In total, eight faces should be black, regardless of how we arrange the small cubes; what matters is only how many faces are colored. Moreover, it doesn't matter if we color, for example, the first and second faces or the first and third faces - we can rotate the small cube to convert one case to another and vice versa. Let's list the possible colorings of the small cubes. The rows record the number of black faces on each of the small cubes, always from the highest number $\mathrm{k}$ to the lowest:

- $3,3,2,0,0,0,0,0$,

$\bullet 3,3,1,1,0,0,0,0$,

- $3,2,2,1,0,0,0,0$,
- $3,2,1,1,1,0,0,0$,
- $3,1,1,1,1,1,0,0$,
- $2,2,2,2,0,0,0,0$,
- $2,2,2,1,1,0,0,0$
- $2,2,1,1,1,1,0,0$,
- 2, 1, 1, 1, 1, 1, 1, 0,
- $1,1,1,1,1,1,1,1$.

This gives us a total of 10 different colorings of the small cubes.