Yet another question about combinations

December 30, 2021 6:13 PM

Consider a square game board made up of four square sub-boards, arranged in a 2x2 grid. Each sub-board can be rotated to one of four orientations. Each sub-board can occupy any one position in the 2x2 grid. All four sub-boards must be used. How many game board combinations are there?

My first instinct was that each sub-board can have one of 16 states - top left, top right, bottom left, bottom right, with one of four orientations. In hexadecimal, that's FFFF, or 65,535 combinations.

My second instinct was that this doesn't account for each sub-board only being able to occupy one position. For example, FFFF can't be a valid board, because this means all four sub-boards are in the same position and orientation - that is, they're stacked rather than adjacent.

Then I thought perhaps the answer is 16 x 15 x 14 x 13 = 43,680, because this is a wild guess and I'm probably confusing a combinatorial thing with a probability thing.

And then my head started to hurt and I came here.

(If you're curious, the board is for Space Crusade, and the claim was that there were "over a thousand combinations", which is starting to seem low, but I'm probably wrong.)

posted by some little punk in a rocket to Science & Nature (5 answers total)

Ignoring orientation, there are 4! (four factorial, or 4 × 3 × 2 × 1 = 24) ways to locate the four sub-boards: You have four choices for the top left, then three choices for the top right, two choices for bottom left, and then just one board left to place in the bottom right.

Once you have chosen their locations, there are 4⁴ (or 4 × 4 × 4 × 4 = 256) ways to orient the sub-boards: 4 ways for the first, 4 for the second, 4 for the third, and 4 for the last.

So the total number of configurations is 4! × 4⁴, or 24 × 256, or a total of 6144 configurations.
posted by mbrubeck at 6:39 PM on December 30, 2021 [3 favorites]

Naively, it's (4x4)x(3x4)x(2x4)x(1x4)=6144.
That's four position choices for the first board (times four rotations), times three remaining choices for the second board position (and four rotations), etc. Each board takes a slot but they don't interfere or disallow rotations of the others.

Now, game-mechanically or aesthetically you may want or need to count certain ones of those as the "same", e.g. if the whole thing is a mirror image of a previous combination, or a rotation, etc. Those are fiddly to deal with but, as written, I think your answer is 6144.
posted by SaltySalticid at 6:40 PM on December 30, 2021 [2 favorites]

Note: If you consider configurations X and Y identical when Y is just X rotated by some amount, then you should divide 6144 by 4 to get the number of unique configurations (since the method above will produce all 4 rotations of each configuration). Then the total is 1536, which fits the claim of “over a thousand combinations.”
posted by mbrubeck at 6:46 PM on December 30, 2021 [5 favorites]

And if the game allows going off one edge and arriving at the opposite edge where the board is essentially a torus you have a few more identical combinations.
posted by zengargoyle at 10:34 PM on December 30, 2021

Came to say the same thing as mbrubeck. It depends on your definitions, of coursej.
posted by number9dream at 11:46 AM on December 31, 2021