Suppose that a robot is placed on the Cartesian plane. At each step, it is…

Concept: For a lattice-path problem where each step increases exactly one of two coordinates by 1, and the total increase required in each coordinate is fixed, the number of distinct monotonic paths from the origin to the target equals the number of ways to arrange the required move-multiset. This is because every such path corresponds to a unique sequence of moves containing a fixed count of each move type, and choosing the positions of one move type within that sequence fixes the entire path — giving the count as a binomial coefficient C(m+n, m), where m and n are the required counts of each move type.

Application:

1. To go from i = 0 to i = 10, the robot must make exactly 10 moves of the form (i, j) → (i + 1, j).

2. To go from j = 0 to j = 10, the robot must make exactly 10 moves of the form (i, j) → (i, j + 1).

3. So every valid path has a fixed total length of 10 + 10 = 20 moves.

4. Each valid path is a distinct arrangement of a sequence of 20 moves containing exactly 10 moves of the first type and 10 of the second type.

5. The number of such arrangements equals the number of ways to choose which 10 of the 20 move-slots are of the first type (the remaining slots are automatically the second type): C(20, 10).

6. Evaluating, C(20, 10) = 20! / (10! × 10!) = 184756.

Cross-check: The formula can be checked on a smaller case: for a robot moving from (0,0) to (1,1), there are exactly two possible move-sequences (first-type-then-second-type, or second-type-then-first-type), so the path count should be 2. The formula gives C(1+1, 1) = C(2,1) = 2, which matches. This confirms the same reasoning correctly counts the paths for the (10,10) case.

Hence, the number of distinct paths from (0,0) to (10,10) is C(20, 10) = 184756.