10

We calculate the number of ways to place two dominos and subtract the number of overlapping pairs
Each domino can be placed horizontally ($1\times 2$) or vertically ($2\times 1$)

• Horizontal placement: Each row has 7 locations to place and each column has 8 locations. So in total, we have $7\times 8=56$ ways.
• Vertical placement: Each column has 7 locations to place and each row has 8 locations. So in total, we have $7\times 8=56$ ways.
  In total we have $56+56=112$ ways to place one domino
  The total number of ways to place two dominos is

$$\left(\genfrac{}{}{0ex}{}{112}{2}\right)=\frac{112\times 111}{2}=6216$$

The number of overlapping pairs can be calculated by categorizing to different conditions

• For corner squares (4 squares), each has 2 ways to cover
• For edge squares (24 squares), each has 3 ways to cover
• For inner squares (36 squares), each has 4 ways to cover
  We have to choose 2 placements in different categories

$$4\times \left(\genfrac{}{}{0ex}{}{2}{2}\right)+24\times \left(\genfrac{}{}{0ex}{}{3}{2}\right)+36\times \left(\genfrac{}{}{0ex}{}{4}{2}\right)=4+72+216=292$$

Then we get the number of non-overlapping ways

$$6216-292=5924$$

And the final answer is

$$5924\mathrm{mod}1000=\boxed{924}$$