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The question is asking which of the given functions has a growth rate that belongs to the Big-O complexity class \( O(2^n) \). Let's analyze the growth rates of the functions in the options:

- **A: \( n! \)**: The factorial function grows faster than exponential functions like \( 2^n \). Therefore, \( n! \) is not \( O(2^n) \).

- **B: \( 3^n \)**: Since \( 3^n \) grows faster than \( 2^n \), \( 3^n \) is not \( O(2^n) \). In fact, \( 2^n \) is \( O(3^n) \).

- **C: \( n^2 \)**: A polynomial function like \( n^2 \) grows much slower than an exponential function like \( 2^n \). Therefore, \( n^2 \) is \( O(2^n) \).

- **D: \( n \log n + 4^n \)**: The term \( 4^n \) grows faster than \( 2^n \) since \( 4^n = (2^2)^n = 2^{2n} \). This implies that \( 4^n \) dominates the function and grows faster than \( 2^n \), so \( n \log n + 4^n \) is not \( O(2^n) \).

### Correct answer: **C: \( n^2 \)**

Would you like further clarification or have more questions on this topic?

Here are five related questions for further exploration:
1. What does Big-O notation represent in terms of algorithm complexity?
2. How does the growth rate of \( n! \) compare to exponential functions like \( 2^n \)?
3. Can you provide examples of other functions that are \( O(2^n) \)?
4. How does the function \( n \log n \) compare in growth to \( 2^n \)?
5. What are common mistakes when comparing growth rates of functions using Big-O notation?

**Tip**: When analyzing Big-O complexity, focus on the dominant term, especially for large \( n \). This term dictates the growth rate of the function.

Select the function that is O(2^n). A) n! B) 3^n C) n^2 D) n log n + 4^n

Analyze different functions and compare their growth rates to O(2^n). Understand the complexity of factorial, exponential, and polynomial functions through a detailed solution to a Big-O notation problem.

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