- #1
Mandelbroth
- 610
- 23
I'm pretty sure I'm right, but I'd appreciate it if I could obtain some verification.
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?
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?