Painting cube with 3 colours (using each colour at least one time) Find no. of ways to colour the cube. (Rotation don't change the colouring.) Solve it for class 9 student in a easier way. Please
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$\begingroup$There are $3^6 $ ways to assign $3$ colours (red, green, blue) to $6$ faces. Subtract the $2^6$ ways that do not use red, $2^6$ ways that do not use green, $2^6$ ways that do not use blue. Rhe $3$ monochromatic way were subtracted twice, so have to be added back in. We count a total of $$ 3^6-3\cdot 2^6+3=540.$$ However, some of these are just rotated versions of another. A rotation of the cibe is either
- a rotation by $90^\circ$ around an axis through two opposing faces. The only 3-colurings invariant under such a rotation are those with these two faces of different colour and the remaining "belt" of four faces of the third colour. There are $18$ such colourings, but we count them only as $3$.
- a rotation by $180^\circ$ around an axis through two opposing faces. The only 3-colurings invariant under such a rotation that are not also $90^\circ$ symmetric are those with opposing faces equal. There are $6$ such colourings, but we count them only as $1$.
- a rotation by $180^\circ$ around an axis through two opposing edges. The only 3-colurings invariant under such a rotation have one pair of opposing faces of one colour, and the other colours used on two adjacent faces each. There are $3\cdot 3\cdot 4=36$ such colourings, but we count them only as $3$.
- a rotation by $120^\circ$ around an axis through two opposing vertices. No 3-colouring can be invariant under such a rotation.
The remaining $540-18-6-36=480$ colourings have no symmetry, hence are identified in groups of $24$ (the total number of cube symmetries) and count only as $20$. Add back the $3+1+3$ symmetric colourings to arrive at a total of$$ 27.$$
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