Let $n$ and $t$ be positive integers. What is the number of ways to place $t$ dominoes $(1\times 2$ or $2\times 1$ rectangles) in a $2\times n$ table so that there is no $2\times 2$ square formed by $2$ dominoes and each $2\times 3$ rectangle either does not have a horizontal domino in the middle and last cell in the first row or does not have a horizontal domino in the first and middle cell in the second row (or both)? 
1. **Define the problem and notation:**
   - We need to find the number of ways to place \( t \) dominoes on a \( 2 \times n \) table such that no \( 2 \times 2 \) square is formed by two dominoes.
   - Let \( A(n, t) \) denote the number of such ways.

2. **Base cases:**
   - \( A(n, 0) = 1 \): There is exactly one way to place 0 dominoes, which is to place none.
   - \( A(n, 1) = 2(n-1) \): There are 2 choices for the row (top or bottom) and \( n-1 \) choices for the position of the left cell of the domino.

3. **Recurrence relation:**
   - Consider the rightmost column of the \( 2 \times n \) table:
     - 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, the recurrence relation is:
   \[
   A(n, t) = A(n-1, t) + 2A(n-2, t-1) + A(n-3, t-2)
   \]

4. **Induction hypothesis:**
   - Assume that \( A(n, t) = \binom{2(n-t)}{t} \) holds for all \( n \) and \( t \).

5. **Induction step:**
   - We need to show that the recurrence relation \( A(n, t) = A(n-1, t) + 2A(n-2, t-1) + A(n-3, t-2) \) is satisfied by \( A(n, t) = \binom{2(n-t)}{t} \).

6. **Verification:**
   - For \( A(n-1, t) \):
     \[
     A(n-1, t) = \binom{2((n-1)-t)}{t} = \binom{2(n-t-1)}{t}
     \]
   - For \( A(n-2, t-1) \):
     \[
     A(n-2, t-1) = \binom{2((n-2)-(t-1))}{t-1} = \binom{2(n-t-1)}{t-1}
     \]
   - For \( A(n-3, t-2) \):
     \[
     A(n-3, t-2) = \binom{2((n-3)-(t-2))}{t-2} = \binom{2(n-t-1)}{t-2}
     \]

   - Using the binomial coefficient identity:
     \[
     \binom{2(n-t)}{t} = \binom{2(n-t-1) + 1}{t} = \binom{2(n-t-1)}{t} + \binom{2(n-t-1)}{t-1} + \binom{2(n-t-1)}{t-2}
     \]

   - This matches the recurrence relation:
     \[
     A(n, t) = A(n-1, t) + 2A(n-2, t-1) + A(n-3, t-2)
     \]

   - Therefore, by induction, \( A(n, t) = \binom{2(n-t)}{t} \) holds for all \( n \) and \( t \).

The final answer is \( \boxed{\binom{2(n-t)}{t}} \)