Number of Paths Using Permutations

[Mandelbroth]

I'm pretty sure I'm right, but I'd appreciate it if I could obtain some verification.

Homework Statement

Consider the following grid:

The goal is to move from point $\alpha$ to point $\beta$ to point $\gamma$ by moving along the edges of the grid from point to point. You can only move to the right or down from $\alpha$ to $\beta$, and you can only move left or up from $\beta$ to $\gamma$. You cannot move outside the grid.

How many distinct paths are there from $\alpha$ to $\beta$ to $\gamma$?

2. The attempt at a solution
I split this into two parts. The first part, $\alpha$ to $\beta$, consists of some combination of moves, but always consists of moving down 4 times and to the right 4 times. Thus, the total number of paths will be $\frac{8!}{4! \cdot 4!} = 70$.

The second part is from $\beta$ to $\gamma$. It's the same thing, but with 2 moves up and 2 moves left. Thus, the number of paths will be $\frac{4!}{2! \cdot 2!} = 6$.

Thus, the total number of paths is $70 \cdot 6 = 420$. Am I correct?

 

[Dick]

I'm pretty sure I'm right, but I'd appreciate it if I could obtain some verification.

Homework Statement

Consider the following grid:

The goal is to move from point $\alpha$ to point $\beta$ to point $\gamma$ by moving along the edges of the grid from point to point. You can only move to the right or down from $\alpha$ to $\beta$, and you can only move left or up from $\beta$ to $\gamma$. You cannot move outside the grid.

How many distinct paths are there from $\alpha$ to $\beta$ to $\gamma$?

2. The attempt at a solution
I split this into two parts. The first part, $\alpha$ to $\beta$, consists of some combination of moves, but always consists of moving down 4 times and to the right 4 times. Thus, the total number of paths will be $\frac{8!}{4! \cdot 4!} = 70$.

The second part is from $\beta$ to $\gamma$. It's the same thing, but with 2 moves up and 2 moves left. Thus, the number of paths will be $\frac{4!}{2! \cdot 2!} = 6$.

Thus, the total number of paths is $70 \cdot 6 = 420$. Am I correct?

Seems fine to me.