The statement `n! gt 2^(n-1), n in N` is true for

Options

A. (a) `n gt 1`
B. (b) `n gt 2`
C. (c) All `n in N`
D. (d) None of these

Answer

To solve the inequality \( n! > 2^{(n-1)} \) for \( n \in \mathbb{N} \), we will check the values of \( n \) starting from 1 and moving upwards to see for which values the inequality holds true.

### Step-by-Step Solution:

1. **Check for \( n = 1 \)**:
- Left-hand side (LHS): \( 1! = 1 \)
- Right-hand side (RHS): \( 2^{(1-1)} = 2^0 = 1 \)
- Comparison: \( 1 \not> 1 \) (not satisfied)

2. **Check for \( n = 2 \)**:
- LHS: \( 2! = 2 \)
- RHS: \( 2^{(2-1)} = 2^1 = 2 \)
- Comparison: \( 2 \not> 2 \) (not satisfied)

3. **Check for \( n = 3 \)**:
- LHS: \( 3! = 6 \)
- RHS: \( 2^{(3-1)} = 2^2 = 4 \)
- Comparison: \( 6 > 4 \) (satisfied)

4. **Check for \( n = 4 \)**:
- LHS: \( 4! = 24 \)
- RHS: \( 2^{(4-1)} = 2^3 = 8 \)
- Comparison: \( 24 > 8 \) (satisfied)

5. **Check for \( n = 5 \)**:
- LHS: \( 5! = 120 \)
- RHS: \( 2^{(5-1)} = 2^4 = 16 \)
- Comparison: \( 120 > 16 \) (satisfied)

6. **General Observation**:
- As \( n \) increases, \( n! \) grows much faster than \( 2^{(n-1)} \). We can observe that for \( n \geq 3 \), the inequality holds true.

### Conclusion:
The inequality \( n! > 2^{(n-1)} \) is satisfied for all \( n \) greater than 2. Therefore, the correct answer is:

**Option B: \( n > 2 \)**.