# How many combinations can be available in a $4\times 4$ grid

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.

• Please explain. When are two combination considered identical? – user228113 May 27 '15 at 10:22

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.
• $\binom{16}{8} = \frac{16!}{8!8!} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$. – N. F. Taussig May 27 '15 at 12:10