1. The number the table holds

Number the rows $0 \ldots m - 1$ from the top and the columns $0 \ldots n - 1$ from the left. Call $P(r, c)$ the number of different routes from the start cell $(0, 0)$ to cell $(r, c)$, moving only down or right. The answer is $P(m - 1, n - 1)$.

2. The value of one cell

The last move into cell $(r, c)$ comes from one of two cells.

The two groups of routes do not overlap, because their last moves differ. Together they are all routes into $(r, c)$. So

$P(r, c) = P(r - 1, c) + P(r, c - 1)$

The first row and the first column are the base cases. A cell in row 0 can only be reached by moving right, so $P(0, c) = 1$. A cell in column 0 can only be reached by moving down, so $P(r, 0) = 1$.

3. Why the table equals a binomial coefficient

Every route from $(0, 0)$ to $(m - 1, n - 1)$ makes exactly $m - 1$ down moves and $n - 1$ right moves. That is $m + n - 2$ moves in total. A route is fixed once you choose which of those moves are the down moves. The number of ways to choose $m - 1$ positions out of $m + n - 2$ is the binomial coefficient:

$P(m - 1, n - 1) = \binom{m + n - 2}{m - 1}$

The recurrence of part 2 is the same rule that builds Pascal's triangle, $\binom{k}{j} = \binom{k - 1}{j - 1} + \binom{k - 1}{j}$. So the table the code fills is Pascal's triangle. The cells with the same value of $r + c$ form one row of the triangle. The code never computes a binomial coefficient. It only adds, and the formula confirms its result.

4. One row is enough

The code keeps a single list row of length $n$ instead of a full $m \times n$ table. At the start, row is the first grid row, all $1$.

For each next grid row, the inner loop runs row[j] += row[j - 1] for $j$ from 1 to $n - 1$. Before this line, row[j] still holds the value of the cell above, $P(r - 1, c)$. row[j - 1] was already updated in this pass, so it holds the cell to the left, $P(r, c - 1)$. The sum is $P(r, c)$.

row[0] is never changed, because column 0 always has $1$ route. The loop must run from left to right. If it ran from right to left, row[j - 1] would still hold the row above.

5. Example 1 by hand

m = 3, n = 7. The list row after each grid row:

The grid shows $P(r, c)$ for every cell of the $3 \times 7$ grid. The three marked cells show the rule of part 2 for the last cell. The cell above it holds $7$, and the cell to its left holds $21$. So the bottom-right cell holds $7 + 21 = 28$.

The code returns row[6] = 28, the expected output. The formula agrees: $\binom{3 + 7 - 2}{3 - 1} = \binom{8}{2} = \frac{8 \cdot 7}{2} = 28$.

6. Example 2 by hand

m = 3, n = 2. The list row starts as [1, 1]. After grid row 1 it is [1, 2], and after grid row 2 it is [1, 3]. The code returns 3, the expected output. The formula agrees: $\binom{3 + 2 - 2}{3 - 1} = \binom{3}{2} = 3$. The three routes are Right, Down, Down; Down, Down, Right; and Down, Right, Down.

7. From the formulas to the code

The two loops run $(m - 1) \cdot (n - 1)$ additions, so the code takes $O(m \cdot n)$ time. The list row takes $O(n)$ space.