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There is a $4\times 4$ grid along with $8$ blue tiles and $8$ white tiles. How many combinations are there to fill the grid? Please explain.

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    $\begingroup$ Please explain. When are two combination considered identical? $\endgroup$ – user228113 May 27 '15 at 10:22
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Once places for all the blue tiles have been chosen, there is no choice about where the white tiles can go. There are 16 places that a blue tile might go and you need to choose 8 of them. There are (by definition) $\binom {16}8$ (16 choose 8) ways to do this.

$\binom{16}8 = \frac{16\times 15\times14\times\cdots\times 9}{8\times7\times6\times\cdots\times1}$ as there arre 16 places for your first tile, 15 for the next, 14 for the one after and so on. We need to divide by $8\times\cdots\times1$ because there are that many different way to choose any set of 8 tiles as in the top part, the order matters but we don't want the order to matter. The reason that there are $8\times7\times\cdots\times1$ reordering is that if you wanted to reorder 8 tiles then there would be 8 places to move the first tile to, 7 for the next, 6 for the one after and so on.

If you count two grids as equivalent if rotation (or reflection) doesn't change them then the problem is more complicated as you need to take into account that only some grids will change if you rotate them so you can't just divide the previous answer by 4.

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  • $\begingroup$ $\binom{16}{8} = \frac{16!}{8!8!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$. $\endgroup$ – N. F. Taussig May 27 '15 at 12:10
  • $\begingroup$ Yes you're right. Edited $\endgroup$ – Dan Robertson May 27 '15 at 12:30

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