What is the maximum number of cuboids with only two face painted one face is painted red and other green?

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Maximum number of cuboid with red colour is possible when cube is painted with red colour in 3 sides with minimum number of common edges (which is equal to 2)
Hence required maximum value is 6 (5 + 5 + 4 - 2) = 72
For minimum number of such cuboid Red colour is used only once and minimum number of cubes in that case is 20
Hence required ratio is 72 : 20 = 18 : 5

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case when k is maximum, one particular colour is used on there faces such that any two faces are adjacent to each other. Required number of cuboids will come from edges but not from vertex = 3 + 4 + 5 + 1 = 13

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
To satisfy this case all the cuboids on the edges and corners must have more than one colour on them. And in that case opposite face must have painted in the same colour.
In that case number of cuboids with 3 colours on them = 8
In that case number of cuboids with 2 colours on them = 4 x (2 + 3 + 4 ) = 36
Hence number of cuboids with at least 1 colour on them is 120 - 36 - 8 = 76

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Number of cuboids with no face painted is (4 - 2)(5 - 2)(6 - 2) = 2 x 3 x 4 = 24

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
No cubes are with 4 face painted.

Total no. of cubes = 5³ = 125,
Some cubes from different corners are removed and the number removed cubes are 2, 3, 4 and 4.
Remaining number of small cubes:
= 125 - 2 - 3 - 4 - 4 = 125 - 13 = 112

In any plane, leave 4 sides cube and select (3 x 3 x 3) inter section. But
the cubes 2 x 2 x 1 give 2 less cube because that part we are already removed.
No. of cubes = (3 x 3 x 3) - 2 = 25.

Only two faces are coloured is when cubes are at the edges (baring the corner cubes)
If no cubes have been removed then on each edges we will get 3 cubes that has exactly 2 faces coloured, hence total number of such cubes = 12 x 3 = 36, because we have 12 edges.
Out of these 3 cubes are removed hence required number of cubes = 36 - 3 = 33

Each has Red faces on top layer = all edges cube = 2 + 2 + 2 + 2 = 8

Number of cubes with 3 face coloured = 4 (Bottom cubes) + 8 top cubes + 4 (column cubes) = 16