jmc

There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation. There are $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins. Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70\cdot 9=\boxed{630}$.