How many of these problems have you posted?

Denote the desired number by $A(n,t)$. If the rightmost column does not contain any domino cells, we have $A(n-1,t)$ possibilities. If the table has a domino in the last two cells of the top row, then below it and below to the left by one cell there are no dominoes (because of the problem condition), so the number of possibilities is $A(n-2,t-1)$. If there is a domino in the last two cells of the bottom row, then either there is no domino on top or top left, which happens in $A(n-2,t-1)$ ways, or there is a domino in the second-to-last and third-to-last cell in the top row (and nothing below it due to the problem condition), which leads to $A(n-3,t-2)$ possibilities. Therefore $A(n,t) = A(n-1,t) + 2A(n-2,t-1) + A(n-3,t-2)$. On th other hand, $A(n,0) = 1$ (we have $1$ way to put $0$ dominoes, by not putting anything) and $A(n,1) = 2(n-1)$ ($2$ choices for a row and $n-1$ for the position of the left cell). Now by induction we obtain $A(n,t) = \binom{2(n-t)}{t}$.