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, and the distance between two vertices can be determined by the edges connecting them. The farthest vertex from any given vertex in a cube is the vertex that is diagonally opposite.

2. **Identifying the Vertices**:
Let's label the vertices of the cube as follows:
- A (0, 0, 0)
- B (1, 1, 0)
- C (1, 0, 0)
- D (0, 1, 0)
- E (0, 0, 1)
- F (1, 1, 1)
- G (1, 0, 1)
- H (0, 1, 1)

Here, A is at (0, 0, 0) and B is at (1, 1, 1), which is the farthest vertex from A.

3. **Calculating the Path**:
To move from A to B, we need to move 1 unit in the x-direction, 1 unit in the y-direction, and 1 unit in the z-direction. This means we need to make a total of 3 moves: one in each direction.

4. **Counting the Distinct Paths**:
The distinct paths can be represented as permutations of the moves. The moves can be represented as:
- X (move in x-direction)
- Y (move in y-direction)
- Z (move in z-direction)

The total number of distinct arrangements of the moves X, Y, and Z is given by the formula for permutations of multiset:
\[ \text{Number of distinct paths} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!} \]
where \( n \) is the total number of moves, and \( n_1, n_2, n_3 \) are the counts of each type of move.

Here, we have:
- Total moves \( n = 3 \) (1 X, 1 Y, 1 Z)
- \( n_X = 1, n_Y = 1, n_Z = 1 \)

Therefore, the number of distinct paths is:
\[ \text{Number of distinct paths} = \frac{3!}{1! \cdot 1! \cdot 1!} = \frac{6}{1} = 6 \]

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