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I How many combinations of unique arrangements are there?

  1. Mar 14, 2016 #1
    If there was a 1 billion x 1 billion x 1 billion cube made of 3D pixel cubes, and half of them are black and half of them are clear/colorless, then how many combinations of unique pixel arrangements are there?

    Would the amount of shapes/objects in this cube be infinite? (Assuming the black pixels represent solids)
     
  2. jcsd
  3. Mar 14, 2016 #2

    phinds

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    You have posited a finite number of things. How would you get from finite to infinite?
     
  4. Mar 15, 2016 #3
    First you could simplify the combinations of pixel cubes to 4x4x4, so 64 pixels total with 32 black and clear. To simplify the cube you could make it a string 64 pixels long that would be folded into a cube after arrangement. Then to find all possible permutations you would do: 64P32=4.822199248906x1053 total possible combinations. This would be bigger with a billion3 pixels.
    This is their total possible combinations as a cube but also as a string, so this is the total possible combinations for any shape and as you could arrange it into an infinite number of shapes in space. The combinations of shape would be infinite, but the permutations would be finite.
     
  5. Mar 16, 2016 #4
    Sorry I got it wrong, for my example of a 4x4x4 it would be 64C32~1.83x1018. So with 1 billion it would be 1 billion cubed choose half one billion cubed. I cant find any calculator to find this value, but it will be the 10000000003/2 th entry on the 10000000003 th line on Pascal's triangle.
     
  6. Mar 16, 2016 #5
    You've used the word "combinations" in a way that isn't the standard usage in mathematics. You should probably just omit the word and say, "How many unique pixel arrangements are there?". Then the answer is just 2^1000000000000000000000000000, which is a humongous number. Unless you did mean a combination, in which case, you should probably reword your question to be more clear.
     
  7. Mar 16, 2016 #6
    For a set of 2N items, the number of subsets of size N is called 2NC2 ("2N choose 2") and is exactly equal to (2N!)/(N!)^2.

    Stirling's approximation to K! is

    K! ~ √(2πK) KK e-K.​

    This gives

    2NC2 ~ √(4πN) (2N)2N e-2N / (√(2πN) NN e-N)2

    which simplifies to

    22N / √(πN).​

    For N = 109, this is approximately 3.7965 × 10602,059,986.

    (Some calculators will say this is infinity.)
     
    Last edited: Mar 16, 2016
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