# Simple combinatorics gone wrong

1. Feb 6, 2013

### x2thay

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

(3264*2448)!*2^24

Meaning, the entire possible positions all the pixels can assume times all the different values each individual pixel can have.

2. Feb 6, 2013

### 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
3. Feb 6, 2013

### Dick

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?

4. Feb 6, 2013

### x2thay

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.

5. Feb 6, 2013

### x2thay

So... (2^24)^(3264*2448) ? Basically 17M^8M? Are you sure that's correct? It seems too simple.

6. Feb 6, 2013

### Dick

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).

7. Feb 6, 2013

### rollingstein

Since jpeg is a lossy format a lot of those combinations ought to evaluate to the same output. I could be wrong.

8. Feb 6, 2013

### x2thay

I should have mentioned earlier, but I meant a bitmap format.

9. Feb 6, 2013

### x2thay

Okay, got it. The solution is a number whose log is 115 805 766.
Thanks, guys.