Just to be consistent here, I'll assume that we're talking about $m$ rows and $n$ columns.
Each path then can be thought of as a Manhattan walk, where we have to go to the right $n-1$ times and down $m-1$ times. The total number of ways to do this is $$\binom{m+n-2}{n-1} = \binom{m+n-2}{m-1}$$ Think of it as either choosing the times we move right (left-hand side) or choosing the times we move down (right-hand side).