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I thought up this logic problem related to the 2048 game. If all 16 tiles on a 2048 board all had the value 1024, how many ways are there to get to the 2048 tile? Here is what I am talking about in an illustration: 2048 board

I found a much simpler, but longer way to think about this: There are 3 ways to combine 2 tiles by going to the right, and 3 by going to the left. That means there are 6 ways to combine the tiles. So, for all the rows and columns, there are $$2 \cdot (4 \cdot 6) = 48$$ ways to get to the 2048 tile.

My question(s) are, is my logic correct? Also, is there a simpler way to approach this logical problem?

Notes

I found two Math.SE post related to 2048 logic, but they have nothing to do with my problem.

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I believe you are Correct.

There are 3 lines separating rows horizontally, 4 pairs of numbers across each line, and 2 ways to combine said pairs (top-down or bottom-up), giving $2*3*4=24$ ways of making pairs.

Repeating for the columns and getting the exact same numbers, we now have $24+24 = 48$ total options for merging two numbers.

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  • $\begingroup$ Okay, that answers the first part. But what about the second prompt. Is my method the simplest way to approach this? Or is there a much more simpler way to approach this? $\endgroup$ – Obinna Nwakwue May 25 '16 at 21:58
  • $\begingroup$ "Is my method the simplest way to approach this?" As far as I can tell, yes. $\endgroup$ – Simpson17866 May 25 '16 at 23:16
  • $\begingroup$ Okay, that helps. Thank you! $\endgroup$ – Obinna Nwakwue May 26 '16 at 0:59
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The corner tiles have 2 ways to move

The side tiles have 3 ways to move

The middle tiles have 4 ways to move

2 3 3 2       2 3 3 3 2       ...
3 4 4 3       3 4 4 4 3
3 4 4 3       3 4 4 4 3
2 3 3 2       3 4 4 4 3
              2 3 3 3 2

Then the formula for different ways to make a $2048$ tile on $(n * n)$ square of $1024$ tiles is:

$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4$

$4 * 2$ (corner tiles)

$(n - 2) * 3 * 4$ (side tiles)

$(n - 2)^2 * 4$ (middle tiles)

$4 * 2 + (n - 2) * 3 * 4 + (n - 2)^2 * 4 =$

$4(n^2 - n)$

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  • $\begingroup$ To complete your "proof", for $n = 4$, $4(n^2 - n) = 4(4^2 - 4) = 4(16 - 4) = 64 - 16 = 48$! $\endgroup$ – Obinna Nwakwue Jun 7 '16 at 15:51

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