Dominoes? More Like Dominate

Plan for counting:
(1)We will first count the number of ways to place one domino on the board.
(2)Next, we choose two of those ways for two dominoes to occupy.
(3)Finally, we subtract out the unwanted cases.

(1) We can position the domino either vertically or horizontally. For each direction we can easily count the number of ways to be $20*5+19*6=214$

(2) We choose $2$ out of the $214$ placements, this gives ${214\choose 2}=22791$ ways to do that.

(3) Note that some of those ways violate our restriction that the two dominoes can't share any squares on the board. Since the $214$ placements are all distinct, the two dominoes can only share one of their $1\times 1$ square in order to form a violation. To count the number of violations,

Case 1: 2 horizontal dominoes
There are $6 \times 18 = 108$ such pairs

Case 2: 2 vertical dominoes
There are \( 20 \times 4 = 188 \( such pairs

Case 3: 1 vertical, 1 horizontal domino
We observe there are \(4\) in a $2\times 2$ square. If we count the number of $2\times 2$ squares in this $6\times 20$ board and then multiply it by $4$ , we will get all the violations without over and under counting. Note that each inner lattice point on the board represents the center point of a $2\times 2$ square, therefore each point determine a square. Since there are $5*19$ inner lattice points, our desired number of violations is $4*5*19=380$

This gives answer $22791-108 - 80- 380=\boxed {22223}$