Question 1210219
Let's analyze this problem carefully. We are looking for paths from A(0,0) to B(3,3) using only steps down (D) or to the right (R), with the added constraint that the path cannot pass through any point more than once (which is automatically satisfied if we only move down or right in a grid).

In a $3 \times 3$ grid, to go from A to B, we need to make a total of 3 steps to the right and 3 steps down, in any order. The total number of steps is $3 + 3 = 6$.

If there were no restrictions other than moving down or right, the number of paths would be given by the number of ways to choose 3 right steps (or 3 down steps) out of the total 6 steps, which is $\binom{6}{3} = \frac{6!}{3!3!} = \frac{720}{6 \times 6} = 20$.

However, the image shows a grid, and it seems the question implies a specific grid configuration where paths might be obstructed or the movement is constrained within the lines of the grid. Assuming the standard interpretation of a grid path from the top-left to the bottom-right using only down and right steps, the constraint of not passing through any point more than once is inherently satisfied.

Let's re-read the question to ensure no details were missed. "Find the number of paths from A to B in the grid below..." The "grid below" is described as 3 by 3. If it's a standard grid, the number of steps required is consistent.

Consider the movement on the grid lines. To go from the top-left intersection to the bottom-right intersection of a $3 \times 3$ grid (meaning 3 units down and 3 units to the right), any path will consist of a sequence of 3 'R' moves and 3 'D' moves. The number of such sequences is $\binom{6}{3} = 20$.

The example path shown in the image confirms this type of movement. The path moves along the grid lines, either down or to the right, at each step.

The constraint "The path cannot pass through any point more than once" is redundant for paths restricted to down and right movements in a grid from the top-left to the bottom-right. Such a path will always be monotonically increasing in both the x and y coordinates, thus never revisiting a point.

Therefore, the problem reduces to finding the number of sequences of 3 'R' moves and 3 'D' moves.

Number of paths = $\binom{\text{total steps}}{\text{number of right steps}} = \binom{6}{3} = 20$.
Alternatively,
Number of paths = $\binom{\text{total steps}}{\text{number of down steps}} = \binom{6}{3} = 20$.

Final Answer: The final answer is $\boxed{20}$