To solve this problem, we need to determine how many different paths can be taken from point $A$ to point $B$ while only moving up and right along the grid.

Understanding the grid:

• The grid starts at point $A$ at $(0,0)$ and ends at point $B$ at $(9,3)$.
• To move from $A$ to $B$, you must make exactly 9 rightward moves (since the horizontal distance is 9 units) and 3 upward moves (since the vertical distance is 3 units).

The number of distinct paths is the number of ways to arrange these 9 rightward moves (R) and 3 upward moves (U) in a sequence of length 12 (since 9 + 3 = 12 total moves).

Key idea:

This is a combinatorics problem where we need to choose 3 positions for the upward moves out of the total 12 moves. This is a "combination" problem, and the number of ways to do this is given by:

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}$

Calculation:

$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220$

Thus, there are 220 distinct paths from $A$ to $B$.

Would you like more details on any part of the solution?

Related questions:

1. How do you calculate combinations for any number of steps?
2. What if there were obstacles in the grid? How would you calculate the number of paths?
3. How does the binomial coefficient apply to other real-world scenarios?
4. What if you could move diagonally as well? How would the number of paths change?
5. How do you generalize this for larger grids?

Tip:

In combinatorics, the order of choosing is crucial: combinations are used when the order doesn't matter, and permutations are used when the order does.