Solution:** The problem can be solved using combinatorics. The robot needs to make a total of \(m + n\) moves: \(m\) moves to the right and \(n\) moves up. The number of distinct paths is equivalent to the number of different ways to arrange these moves, which is given by the binomial coefficient:

["Solving Robot Path Problems with Combinatorics: Counting Distinct Movements Using Binomial Coefficients", "When programming a robot to navigate a grid, one common challenge is determining the number of distinct paths from a starting point to a destination using only right and up moves. This seemingly simple task is a classic application of combinatorics—and more precisely, binomial coefficients.", "### The Grid Path Problem Explained", "Consider a robot confined to a 2D grid starting at the origin point (0, 0) and needing to reach the point (m, n). The only allowed moves are:\n- Moving right (m) times\n- Moving up (n) times", "Regardless of the order, as long as the robot performs exactly (m) right moves and (n) up moves, it will arrive at the destination. The core question becomes: How many unique sequences of moves can lead the robot from (0, 0) to (m, n)?", "### Why Combinatorics Works Here", "Each valid path is a permutation of a sequence containing (m) right moves (e.g., R, R, ..., R) and (n) up moves (e.g., U, U, ..., U), totaling (m + n) moves. Since multiple right moves are indistinguishable from each other, and likewise for up moves, the number of distinct arrangements is not ( (m+n)! )—which counts all permutations independently—but the number of ways to choose positions for the right moves (or equivalently, the up moves) within the full sequence.", "This is precisely what the binomial coefficient captures:", "[\n\boxed{\binom{m+n}{m} = \frac{(m+n)!}{m! \cdot n!}}\n]", "Or equivalently,", "[\n\binom{m+n}{n} = \frac{(m+n)!}{n! \cdot m!}\n]", "Both expressions yield the same result—the number of unique combinations of paths.", "### What This Means in Practice", "- Each binomial coefficient calculates the number of distinct sequences with exactly (m) right moves and (n) up moves.\n- Instead of listing paths one by one—which is computationally inefficient—we use combinatorics to compute the total number directly.\n- This approach scales efficiently even for large (m) and (n), making it ideal for real-time path planning or simulation.", "### Example", "Suppose (m = 3) (3 right moves) and (n = 2) (2 up moves). A valid path might be RRUU, RURU, RUUR, etc. The total number of distinct paths is:", "[\n\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10\n]", "Hence, there are 10 unique ways the robot can reach its destination using exactly 3 right and 2 up moves.", "### Summary", "Using combinatorics and the binomial coefficient, we solve the robot path problem elegantly and efficiently. This method transforms an exponential enumeration challenge into a simple, scalable calculation—showcasing how powerful mathematical tools empower smart pathfinding and algorithmic design.", "For developers and roboticists, leveraging combinatorial logic streamlines path validation, animation scripting, and resource-aware motion planning—making it an essential concept in automation and computational geometry."]