Let A and B be the vertices of cube such that B is farthest away A the number of the distinct paths from A to B along the edges of cube is ________________________

Answer

To find the number of distinct paths from vertex A to vertex B in a cube where B is the farthest vertex from A, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Cube Structure**:
- A cube has 8 vertices. Let's label them as follows:
- A (0, 0, 0)
- B (1, 1, 1)
- C (1, 0, 0)
- D (0, 1, 0)
- E (0, 0, 1)
- F (1, 1, 0)
- G (1, 0, 1)
- H (0, 1, 1)
- In this case, A is at (0, 0, 0) and B is at (1, 1, 1), which are diagonally opposite vertices.

2. **Identifying the Path**:
- To move from A to B, we need to move in the positive x, y, and z directions. Each move corresponds to traversing along the edges of the cube.

3. **Counting Moves**:
- To go from A to B, we need to make 3 moves: one in the x-direction, one in the y-direction, and one in the z-direction. The order of these moves can vary.

4. **Calculating Distinct Paths**:
- The problem can be reduced to finding the number of distinct arrangements of the moves. Since we have 3 moves (x, y, z), we can think of this as arranging the letters "XYZ".
- The number of distinct arrangements of these letters is given by the formula for permutations of multiset:
\[ \text{Number of distinct paths} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!} \]
- In our case, we have:
- n = 3 (total moves)
- n1 = 1 (for x)
- n2 = 1 (for y)
- n3 = 1 (for z)
- Therefore, the number of distinct paths is:
\[ \text{Number of distinct paths} = \frac{3!}{1! \cdot 1! \cdot 1!} = 6 \]

5. **Conclusion**:
- The total number of distinct paths from vertex A to vertex B along the edges of the cube is **6**.