Suppose there is a cube and we can colour the cube's faces

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When coloring a cube's six faces with two colors, black and white, the initial calculation suggests 64 combinations. However, due to symmetry and indistinguishable patterns, the actual number of unique colorings is reduced to 10. This includes variations for all-black, all-white, and combinations of black and white faces, considering adjacent and non-adjacent placements. The discussion highlights the application of Burnside's theorem for counting these patterns, although there is some confusion regarding Polya's theorem and its derivation. Overall, the key takeaway is that the unique color patterns for the cube total 10.
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suppose there is a cube and we can colour the cube's faces with only two colours ..i.e.
black and white ,,how many different patterns are possible...
 
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Seems like the perfect place to apply Burnside's theorem (about orbits).
 
Here's my take:

How many faces are on a cube? 6
How many allowable variations per face? 2
Combinations possible: 2*2*2*2*2*2 =64

Oh, but wait, this is not a strictly linear situation!
My mind is melting.
 
all white: 1 version
1 black side: 1 version
2 black: 2 versions (adjacent/nonadjacent)
3 black: 2 versions (three in a row/three sides with common vertex)
and by symmetry...
4 black = 2 white: 2 versions
5 black: 1 version
6 black: 1 version

Total... 10 indistinguishable ways.
 
Thanks Rach,,
i am clear with how that value 10 comes..but actually the theory related with Polya's theorem is n't much clear to me...i am actually not clear with how they have derived the formula.
 
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