Let `P(n): 2^(n) lt n!`, then the smallest positive integer for which P(n) is true is:
(i) 1
(ii) 2
(iii) 3
(iv) 4

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the smallest positive integer \( n \) for which the inequality \( 2^n < n! \) holds true. We will check each of the given options step by step.

### Step-by-Step Solution:

1. **Check for \( n = 1 \)**:
- Calculate \( 2^1 \) and \( 1! \):
\[ 2^1 = 2 \]
\[ 1! = 1 \]
- Check the inequality:
\[ 2 < 1 \quad \text{(False)} \]

2. **Check for \( n = 2 \)**:
- Calculate \( 2^2 \) and \( 2! \):
\[ 2^2 = 4 \]
\[ 2! = 2 \]
- Check the inequality:
\[ 4 < 2 \quad \text{(False)} \]

3. **Check for \( n = 3 \)**:
- Calculate \( 2^3 \) and \( 3! \):
\[ 2^3 = 8 \]
\[ 3! = 3 \times 2 \times 1 = 6 \]
- Check the inequality:
\[ 8 < 6 \quad \text{(False)} \]

4. **Check for \( n = 4 \)**:
- Calculate \( 2^4 \) and \( 4! \):
\[ 2^4 = 16 \]
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
- Check the inequality:
\[ 16 < 24 \quad \text{(True)} \]

Since the inequality \( 2^n < n! \) holds true for \( n = 4 \) and not for any smaller positive integers, we conclude that the smallest positive integer for which \( P(n) \) is true is:

\[ \boxed{4} \]