Intermediate
Discrete Mathematics & Math
Q62 / 100
What is Stirling's approximation?
Correct! Well done.
Incorrect.
The correct answer is A) n! ≈ n^n * e^(-n) * √(2πn)
A
Correct Answer
n! ≈ n^n * e^(-n) * √(2πn)
Explanation
Stirling's approximation (more precisely option d): n! ≈ √(2πn) * (n/e)^n. Option a is also correct notation. Used in algorithm analysis for log(n!) = Θ(n log n) and combinatorics. Proves information-theoretic Ω(n log n) sorting lower bound.
Progress
62/100