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Homework Help: Simple combinatorics gone wrong

  1. Feb 6, 2013 #1
    1. So consider an 8 megapixel picture (res: 3264x2448).
    Now, it seems rather simple but I just can't figure out how to calculate the entire number of possible shots/photographs one can take within that resolution, assuming each pixel can have 16777216 different values/colors.

    2. Relevant equations

    3. The attempt at a solution
    So I realize the number has to be absolutely monstruous, so here's what I've tried so far


    Meaning, the entire possible positions all the pixels can assume times all the different values each individual pixel can have.
  2. jcsd
  3. Feb 6, 2013 #2


    Staff: Mentor

    It seems your solution is correct for a JPG image but for a GIF you limited to 256 colors per image. Each color may range from 0 to 16million as a GIF uses an 8 bit value that indicates which color register from the color palette to use.

    One problem I see is that you'll get shots all in one color or half one color half another...

    Do you want to define what a photo is? like it must have a minimum of X colors per photo?

    Next what about color changes that are imperceptibly small? Do you want to instead say that while RED can range from 0 to 255 in value we'll limit it down to a subset of {0, 16, 32, 48, 64 ... 255} in other words 64 color choices.
    Last edited: Feb 6, 2013
  4. Feb 6, 2013 #3


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    If you have 2^24 possible values of each pixel, then for two pixels you would have 2^24*2^24 possibilities, right? What about 3264*2448 pixels?
  5. Feb 6, 2013 #4
    No matter how insignificant the difference is, I meant to calculate e-v-e-r-y single possible matrix 3264x2448 arrangement, given that each entry can assume 2^24 different values. so yes, there will be an enormous amount of shots in which the only difference from the next, will be a single pixel.
  6. Feb 6, 2013 #5
    So... (2^24)^(3264*2448) ? Basically 17M^8M? Are you sure that's correct? It seems too simple.
  7. Feb 6, 2013 #6


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    Yes, I'm sure. It's actually too simple for me to be wrong. It's (number of possibilities for each choice)^(number of choices).
  8. Feb 6, 2013 #7


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    Gold Member

    What format? You didn't answer.

    Since jpeg is a lossy format a lot of those combinations ought to evaluate to the same output. I could be wrong.
  9. Feb 6, 2013 #8
    I should have mentioned earlier, but I meant a bitmap format.
  10. Feb 6, 2013 #9
    Okay, got it. The solution is a number whose log is 115 805 766.
    Thanks, guys.
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