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

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Discussion Overview

The discussion revolves around the problem of coloring the faces of a cube using two colors, black and white. Participants explore the number of distinct patterns that can be formed, considering the effects of symmetry and combinatorial principles.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes that there are 64 combinations of colors if each face can be independently colored black or white.
  • Another participant suggests that the situation is more complex due to symmetry, indicating that not all combinations are unique.
  • A different participant outlines a breakdown of patterns based on the number of black faces, arriving at a total of 10 indistinguishable ways to color the cube.
  • One participant expresses confusion regarding the application of Polya's theorem and its derivation, indicating a lack of clarity on the theoretical background.

Areas of Agreement / Disagreement

There is no consensus on the total number of distinct patterns, as participants present differing views on the impact of symmetry and the application of combinatorial theorems.

Contextual Notes

Participants mention the need for symmetry considerations and the application of Burnside's and Polya's theorems, but do not fully resolve how these theories apply to the problem at hand.

heman
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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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