Gorgeous grids

For the first square, we have $8$ options. For the two neighbors we have $7$ choices for each, so we have $49$ pairs of the two neighbors of the first color. What happens with the last color? There are two cases:

Case 1: The neighbors are different. This is possible in $7*6 = 42$ ways. In this case, as both are different, for the last square we have $6$ options. So completing this case we have $8\cdot 42\cdot 6 = 2016$ .

Case 2: The neighbors are the same. This is possible in $7$ ways. Similarly, $8\cdot 7\cdot 7 = 392$ .

So in total we have $2408$ different boards ignoring the condition of the rotation .

Now let's discover which ones are repeated. Note that every board that we could construct, excepting one type that I'll show below, takes a different color order, or orientation, when you rotate it the $4$ times. The special case is the arrangement of the form:

$\begin{matrix} a & b \\ b & a \end{matrix}$

It's clear enough that there are $56$ of those cases. Also, they have only $2$ different orientations. Now it's convenient to separate $2408$ into $2352 + 56$ . For the other $2352$ there are $4$ possible orientations, and for the $56$ , two orientations. This takes us to the answer which is $\frac {2352}{4} + \frac {56}{2} = 588 + 28 = \boxed {616}$ .

• If all four colors are different, there are ${8 \choose 4}$ ways to choose the colors, and $3! = 6$ ways to order them.

• If there are three different colors, then we must have one color twice on the diagonal, and the other two colors in the other two spots. There are ${8 \choose 1}{7 \choose 2}$ ways to choose the colors and just $1$ way to order them.

• If there are just two different colors, then we must have one color on one diagonal, and one on the other. There are ${8 \choose 2}$ ways to choose two colors and just $1$ way to order them.

The answer is therefore ${8 \choose 4} \cdot 6 + {8 \choose 1}{7 \choose 2} \cdot 1 + {8 \choose 2} \cdot 1 = \boxed{616}$