Call the rows A, B and C. The vertical dominoes (V) must both be placed across A and B or B and C, otherwise there will be gaps. Consider just the two rows containing the vertical dominoes. They must be placed so that they are separated by an even number of grid cells, otherwise it will not be possible to slot in the horizontal dominoes (H). If the grid cells are numbered 1 to 52, then we can reference each cell by its row and cell number. Take the two rows to be A and B. Place a V across A1 and B1 and the other V across A52 and B52. The H can start with the first H at A2-A3 and the 25th H at A50-A51. The 26th H starts at B2-B3 and the 50th at B50-B51. This is arrangement 1.
Move the second V across A50-B50 leaving space for H at A51-A52 and B51-B52. This is arrangement 2.
Arrangement 3 is the 2nd V at A48-B48 leaving space for 4 H at A49-A52 and B49-B52.
Arrangement 26 is the V at A2-B2 leaving space for all the H in A3-A52 and B3-B52.
Arrangement 27 is V at A3-B3 and the other V at A52-B52. But now there are only 25 places where the second V can fit.
So we have a series: 26+25+24+...+1=26*27/2=351.
So in two rows there are 351 different domino arrangements.
What about the third row? This contains H only and the row can be A with rows B and C containing the V; or the row can be C with A and B containing the V. Because there are two possible alternatives there are 2*351=702 ways of arranging the dominoes.