In the worst case, the smallest number of comparisons required by any sorting algorithm, which does only binary comparisons, is

Explanation

A comparison-based sort must be able to distinguish between all n! possible orderings of the input, and each binary comparison can split the remaining possibilities by at most a factor of 2. This information-theoretic argument requires a decision tree of depth at least log₂(n!), so the worst case requires at least ⌈log₂(n!)⌉ comparisons — the well-known lower bound that algorithms like merge sort asymptotically achieve. Hence the correct answer is ⌈log n!⌉.