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how many ways can a standard 6-sided die be painted such that no 2 faces sharing an edge have the same color​

Sagot :

There are 6 faces on a standard 6-sided die. For the first face, there are 6 possible colors to choose from. For the second face, there are 5 remaining colors to choose from. For the third face, there are 4 remaining colors to choose from, and so on. So, the total number of ways to paint the die such that no two faces sharing an edge have the same color is given by the permutation formula:

[tex]$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$[/tex]

So, there are ways this many ways to paint a standard 6-sided die in this manner:

[tex]$\boxed{720}$[/tex]

That's an interesting math problem! Let's think about it.

A standard 6-sided die has 6 faces, each with a different number of dots on it. If we paint the faces with different colors, we want to make sure that no two faces sharing an edge have the same color.

We can represent the faces of the die with the numbers 1 through 6. If we paint the faces in such a way that no two adjacent faces have the same color, then we need to find a way to assign each face a different color.