To solve the problem of finding the number of paths from point $A(0, 0)$ to point $B(6, 3)$ on a Cartesian grid, where you can only move right or up, we can use combinatorics.

Explanation:

Each path from $A$ to $B$ involves exactly 6 moves to the right (R) and 3 moves up (U). Therefore, the total number of moves is $6 + 3 = 9$ moves.

Out of these 9 moves, we need to choose 3 moves to be "up" moves (or equivalently, 6 moves to be "right" moves). The number of such combinations is given by the binomial coefficient $\binom{9}{3}$.

Calculation:

The binomial coefficient $\binom{n}{k}$ is calculated using the formula:

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

Here, $n = 9$ and $k = 3$, so:

$\binom{9}{3} = \frac{9!}{3! \cdot (9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$

Conclusion:

Thus, the number of paths from $A(0, 0)$ to $B(6, 3)$ is 84.

Would you like more details or have any other questions?

Here are 5 related questions:

1. How many paths are there from $A(0, 0)$ to $C(4, 2)$ under the same conditions?
2. What if you could move in all four directions (up, down, left, right) but still need to end at $B(6, 3)$?
3. How does the number of paths change if obstacles are placed on certain grid points?
4. Can you generalize the number of paths for any destination $(m, n)$?
5. How would you compute the number of paths if diagonal movements were allowed?

Tip: The binomial coefficient is a fundamental concept in combinatorics and is widely used in problems involving paths, permutations, and probability.