91 of 100: Cube Color Conundrum

This should not be 5/5 cacti. Way too easy. Just literally count all the possibilities.
0 blue faces
1 blue face
2 adjacent blue faces
2 opposite blue faces
3 blue faces around a vertex
3 blue faces in a line
4 blue faces with 2 adjacent red faces
4 blue, 2 opposite red faces
5 blue faces
6 blue faces
The more formal solution would use Burnside's Counting Lemma , but this was so easy, just literally counting them the fastest thing to do.

It should be immediately clear that you really only need to solve for four cases, since the solution for 0 red, 1 red and 2 red is the same as the solution for 0 blue, 1 blue and 2 blue.

So the solutions are as follows:

• 0 red (6 blue): 1 solution

• 1 red (5 blue): 1 solution

• 2 red (4 blue): 2 solutions (adjacent & non-adjacent)

• 3 red (3 blue): 2 solutions (touching-a-common-corner & in-a-row)

• 4 red (2 blue): 2 solutions (same as 2 red, except now blue are adjacent & non-adjacent)

• 5 red (1 blue): 1 solution (same as 1 red)

• 6 red (0 blue): 1 solution (same as 0 red)

I think these images cover the 10 general cases that I (and everyone else here) found. Please let me know if I made any errors!

• 6 red, 0 blue: 1 way
• 5 red, 1 blue: 1 way
• 4 red, 2 blue: 2 ways (blue sides either bound to each other or opposite each other)
• 3 red, 3 blue: 2 ways (either both colors in one line or both colors meet in one vertex)
• 2 red, 4 blue: 2 ways (red sides either bound to each other or opposite each other)
• 1 red, 5 blue: 1 way
• 0 red, 6 blue: 1 way

The total is 10 ways.

Take a red colour as 1 and blue colour as zero

So to colour six phases, let the pattern be something like - RRRBBB

R- RED B-BLUE

So seven combinations to colour six sides of cube are-

111111-in this only one way to colour

111110-in this only one way to colour

111100-in this two ways to colour--opposite sides blue coloured and adjacent sides blue coloured

111000-in this two ways to colour--3 adjacent sides in plane and 2 in plane and one top or bottom

110000-in this two ways to colour--opposite sides red coloured and adjacent sides red coloured

100000-in this only one way to colour

000000-in this only one way to colour

It can be thought of as 6C2, analogous to how many ways are there to picking a group of 2 people from 6. Chris

i think the correct answer wd be 9. ... if each side must be coloured (which is clearly mentioned in the qs).

I only counted the scenarios from red's perspective and just inverted it where appropriate.

This is the way I thought about it:

All Red inv. All Blue

1 Red inv. 1 Blue

2 adjacent Red inv. 2 adjacent Blue

2 opposite Red inv. 2 opposite Blue

3 Red sharing vertex ( No Inversion)

3 Red in a line ( No Inversion)

Hence, if you count it all up, then there are 10 unique ways.

I have mentally counted just nine colorings.

The smallest answer is ten, greater than nine.

So I have picked ten as the closest.