A man has to move 9 steps. He can move in 4 directions: left, right, forward, backward in how many ways he can move 9 steps such that he finishes his journey one step away (either left orright or forward or backward) from the starting position?

Options

A. `(""^(10)C_(5))^(2)`
B. `4xx(""^(9)C_(4))^(2)`
C. `(""^(9)C_(4))^(2)`
D. `4xx(""^(10)C_(5))^(2)`

Answer

To solve the problem of how many ways a man can move 9 steps such that he finishes his journey one step away from the starting position, we can break down the problem systematically.

### Step-by-Step Solution:

1. **Understanding the Directions**:
The man can move in four directions: left (L), right (R), forward (F), and backward (B). We need to ensure that after 9 steps, he ends up one step away from the starting position.

2. **Defining the Conditions**:
To be one step away from the starting position, the man can end up in one of the following configurations:
- 1 step left (L) and 8 steps in other directions.
- 1 step right (R) and 8 steps in other directions.
- 1 step forward (F) and 8 steps in other directions.
- 1 step backward (B) and 8 steps in other directions.

3. **Calculating Each Configuration**:
We will calculate the number of ways for each configuration separately.

**Case 1**: 1 step left (L) and 8 steps (which can be R, F, B).
- Total steps = 9
- Steps configuration: 1 L, x R, y F, z B such that x + y + z = 8.
- The total number of ways to arrange these steps is given by:
\[ \text{Ways} = \frac{9!}{1! \cdot x! \cdot y! \cdot z!} \]
- The number of non-negative integer solutions to \(x + y + z = 8\) is given by the stars and bars method:
\[ \text{Number of solutions} = \binom{8 + 3 - 1}{3 - 1} = \binom{10}{2} = 45 \]
- Therefore, the number of ways for this case is:
\[ 9! \cdot 45 \]

**Case 2**: 1 step right (R) and 8 steps (which can be L, F, B).
- This case is symmetric to Case 1, so it also contributes the same number of ways:
\[ 9! \cdot 45 \]

**Case 3**: 1 step forward (F) and 8 steps (which can be L, R, B).
- Similar to the previous cases, this also contributes:
\[ 9! \cdot 45 \]

**Case 4**: 1 step backward (B) and 8 steps (which can be L, R, F).
- Again, this is symmetric and contributes:
\[ 9! \cdot 45 \]

4. **Total Ways Calculation**:
Now, we sum all the contributions from the four cases:
\[ \text{Total Ways} = 4 \times (9! \cdot 45) \]

5. **Final Calculation**:
- Calculate \(9!\):
\[ 9! = 362880 \]
- Therefore:
\[ \text{Total Ways} = 4 \times (362880 \cdot 45) = 4 \times 16329600 = 65222400 \]

### Conclusion:
The total number of ways the man can move 9 steps and end up one step away from the starting position is **65,522,400**.