Question 1210221
You're asking for the number of paths from the top-left corner (A) to the bottom-right corner (B) of a 3x3 grid, where each step is either down or to the right.

In a 3x3 grid, to get from A to B, you need to make a total of:
* 3 steps to the right (R)
* 3 steps down (D)

Any valid path will consist of a sequence of these 6 steps (3 R's and 3 D's). The problem then becomes finding the number of distinct arrangements of these 6 steps.

This is a combination problem. We have 6 total steps, and we need to choose 3 of them to be "right" steps (the remaining will be "down" steps). The number of ways to do this is given by the binomial coefficient:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

Where:
* $n$ is the total number of steps (6)
* $k$ is the number of steps in one direction (either 3 right or 3 down)

Let's calculate this:

$$ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (3 \times 2 \times 1)} = \frac{720}{6 \times 6} = \frac{720}{36} = 20 $$

Therefore, there are **20** different paths from A to B in the 3x3 grid following the given rules.

**Answer:** There are **20** paths from A to B.