How many possible paths

Question

Consider the point $\vec{p} = (3,48,4,5,8) \in\mathbb{R}^{5}$, how many distinct paths are there from zero to $\vec{p}$ if the only movements allowed are unit movements in the directions $\vec{e}_{1}, \vec{e}_{2}, \vec{e}_{3}, \vec{e}_{4}, \vec{e}_{5}$?

Would this mean that we have $\frac{(3+48+4+5+8)!}{3!48!4!5!8!} = \frac{68!}{3!48!4!5!8!}$ distinct paths?

Answer 1

Your answer $\frac{(3+48+4+5+8)!}{3!48!4!5!8!} = \frac{68!}{3!48!4!5!8!}$ is exactly correct.

We have to 'line up' 68 moves, so that would suggest $68!$ possibilities, but since you have $3$ $e_1$-moves, they are indistinguishable in this line-up, so the $68!$ overcounts by a factor of $3!$, mneaning that you need to divide $68!$ by $3!$. But the same is true for the other dimensions.

Answer 2

Assuming "up and right" means we are only allowed to move in the direction of $\vec{e}_1, \vec{e}_2, \vec{e}_3, \vec{e}_4$ and $\vec{e}_5$, not the linear combinations of them and not scaled versions of them, I think this is a simple combinatorics question. Equivalent problem would be "Given five letters A, B, C, D and E, how many 68 letters words can you make such that there are 3 As, 48 Bs, 4 Cs, 5 Ds and 8 Es?"

Answer 3

You need to pick $3+48+4+5+8=68$ moves, so you want to know how many ways you can order those. However, there are five move types, corresponding to the directions, and the order in which the moves of a single type are picked does not count. Can you take it from there?