0
$\begingroup$

I am trying to figure out a permutation question. It is related to an art project I am doing with recombinable panels.

I have a grid of $64$ panels in an $8\times8$ arrangement. The panels can be placed in any square in the grid, so $64$ possible places for $64$ possible panels. So $64\times64=4096$ possible arrangements so far.

However the panels each have 2 states “up” or “down”. Does this mean I multiply the $4096$ by $2$, or by another $64$? My logic being I multiply by $2$ as I am effectively making $128$ source panels for the grid of $64$, since an individual panel can not be in the grid in each state simultaneously, as opposed to multiplying by another $64$ which would be valid if a panel could be in the grid in both states simultaneously. i.e. Panel A can be in Grid Square 1 but nowhere else in the grid, and Panel A can be "up" or "down" but not both at the same time. This is true for all other 63 panels at the same time to compose one permutation of the 64 panel grid.

However, once I have that number ($\#$ of possible permutations for the $64$ panel grid), I think I multiply it by $9$, as the entire grid of $64$ panels can be recombined with itself in a meta-grid of $9$ meta-panels ($3\times3$).

So ultimately I am also trying to figure out how many permutations of the $3 \times 3$ meta-grid the 64 panel set could generate, considering their two states.

Any help sorting that out?

  • 0
    There are 64! total ways to place the panels - and that is a huge number (in comparison to 4096 anyway) – 2017-01-12
  • 0
    $2^{n}$ ways to place $n$ panels up or down. $2^{64!}$ is even bigger – 2017-01-12

1 Answers 1

1

Problem 1) If you have $n$ places to put $n$ things, then you have $n!$ ways of doing it.

proof: the first one has a choice of $n$, the second a choice of $n-1$ and so on. You multiply all the choices to get the total amount, so $n!$

$64!$ is big.

Problem 2) If you have $n$ things to be placed upside up (?) or upside down, then there are $2^n$ ways to do it.

proof: First can be done up or down, so 2 choices, then next has 2 and so on. So all $n$ give 2 choices. As before we product, and you get $2^n$.

Now, you haven't said much about the tiles, else we might be able to cancel out any which are the same permutation. But $2^{64!}$ is very big.

  • 0
    Wow, thanks! much larger than I'd figured. The panels are each unique, and can not be in the arrangement more than once, though that instance can be "up" or "down" for each panel individually. So ultimately it sounds like just the grid of 64 can generate a massive amount of permutations, so extrapolating that to a meta-grid of 3X3 64-panel sets makes a number of possibilities so ludicrously large so as to be damn near endless! – 2017-01-13
  • 0
    I know this was a while ago, but if I answered your question(s), consider letting me be the correct answer. It lets me get some points (I have a hard question, and I want as many points to set as high a bounty on it) – 2017-03-05