The trick to making these problems intuitive is to mentally rewrite them into a "how many permutations of this string are possible?" problem. Consider the 2 * 3 case.

    ...
    ...

One way you might get there is

    Right, right, right, down, down

Then you can rewrite this as

    RRRDD

You will always need 3 R's, and you will always need 2 D's. So how many unique strings can be made with this?

Well let's actually consider the degenerate cases.

    ABCDE

there are 5 places A can go, then 4 left B can go, then 3 left C can go, and so on, until we get 5! = 120 possible permutations of ABCDE. If you replace the B with another A to get

    AACDE

now there are only 60 permutations, because half of the original 120 only differed by where the A and the B were relative to one another. By that same logic,

    AACCE

has only 30 combinations, and

    AACCC

has only 10 (seeing why it's 10 and not 20 is actually the trickiest part imo, it's because there are 3! ways to arrange CDE, but only 1 to arrange CCC).

AACCC is isomorphic to RRDDD, which is how we get 10 possible paths to solve the 2*3 grid. We can check this with the binomial theorem: ((2+3) choose 3) = 10.

What's nice about this step by step approach is that it generalizes not just to non-square grids, but to multiple dimensions as well! Imagine trying to get from the top of a 3 by 3 by 3 Rubik's cube to the bottom, how do you do that? Well how many ways are there to rewrite

    AAABBBCCC

? The logic above would suggest 9! / (3! 3! 3!) = 1,680 unique paths. And you can just derive it by starting from the degenerate case and figuring out how to slice things up!