jmc

The most obvious way to solve this problem would be to list Pascal's Triangle to Row 8.

$$\begin{array}{rccccccccccccccccc}Row0:& & & & & & & & & 1& & & & & & & & \\ \text{noalign}\text{smallskip}\text{smallskip}Row1:& & & & & & & & 1& & 1& & & & & & & \\ \text{noalign}\text{smallskip}\text{smallskip}Row2:& & & & & & & 1& & 2& & 1& & & & & & \\ \text{noalign}\text{smallskip}\text{smallskip}Row3:& & & & & & 1& & 3& & 3& & 1& & & & & \\ \text{noalign}\text{smallskip}\text{smallskip}Row4:& & & & & 1& & 4& & 6& & 4& & 1& & & & \\ \text{noalign}\text{smallskip}\text{smallskip}Row5:& & & & 1& & 5& & 10& & 10& & 5& & 1& & & \\ \text{noalign}\text{smallskip}\text{smallskip}Row6:& & & 1& & 6& & 15& & 20& & 15& & 6& & 1& & \\ \text{noalign}\text{smallskip}\text{smallskip}Row7:& & 1& & 7& & 21& & 35& & 35& & 21& & 7& & 1& \\ \text{noalign}\text{smallskip}\text{smallskip}Row8:& 1& & 8& & 28& & 56& & 70& & 56& & 28& & 8& & 1\\ \text{noalign}\text{smallskip}\text{smallskip}& & & & & & & & & & & & & & & & & \end{array}$$

We then add $1+8+28+56+70+56+28+8+1=\boxed{256}.$

An interesting note: We can sum the numbers in some of the smaller rows of Pascal's Triangle. Row 0 sums to $1,$ Row 1 sums to $1+1=2,$ Row 2 sums to $1+2+1=4,$ Row 3 sums to $1+3+3+1=8,$ and Row 4 sums to $1+4+6+4+1=16.$ We start to notice a pattern: the sum of the numbers in Row $n$ of Pascal's Triangle equals $2^n.$ Sure enough, the sum of the numbers in Row 8 is $256,$ which is $2^8.$