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A comparison sort learns the ordering of sort keys only by comparing pairs. The keys are otherwise opaque: their numeric values, digits or encodings cannot choose an array index or a branch. It may move complete sorting records and retain previous answers. Merge sort and the adjacent-swap insertion sort are comparison sorts; counting sort and radix sort exploit additional information.

Any deterministic comparison sort correct for all orders of $n$ distinct sort keys needs at least $\lceil {\log }_2(n!)\rceil$ comparisons on some input. Here $n!$ is a factorial, ${\log }_2$ a logarithm, and the brackets denote the ceiling function. This is a worst-case requirement, not a requirement on every input.^[ref]

For three distinct sort keys, the diagram has one leaf for each possible ordering of the original entries $a,b,c$. Some paths need two comparisons, others three; two comparisons cannot distinguish all six orders.

For the proof outline, represent adaptive comparisons by a binary tree: distinct sort keys give only two outcomes per comparison. A leaf fixes an output permutation; one leaf cannot correctly handle two different strict orders. Hence there must be at least $n!$ leaves. Maximum depth $h$ permits at most $2^h$ leaves, so $2^h\ge n!$ gives the bound. Even a three-way comparator has only two reachable outcomes on this distinct-key family.

The growth follows without a precise approximation to the factorial. At least $\lfloor n/2\rfloor$ of its factors are at least $n/2$, so for $n\ge 2$,The brackets denote the floor function. This gives a constant times $n\log n$ lower bound for large $n$, matching the worst-case comparison growth of merge sort.

Distinct sort keys are essential to the $n!$ counting argument. Identical sort keys do not create $n!$ different strict orders. If all sort keys coincide, stable sorting returns the arrival order. This does not remove the worst-case bound for a comparison sort required to handle arbitrary distinct-key inputs.

Radix sort sorts encoded sort keys a digit at a time. Its least-significant-digit version uses stable sorting on each successive digit, often by counting sort. For $d$ digits with $R$ possible values each, it takes $O(d(n+R))$ time complexity in Big O notation, assuming constant-cost digit extraction and indexing. Stable sorting retains the ordering established by earlier digits when a later digit ties. Both digit access and the direct indexing in counting sort lie outside the comparison sort model, so these bounds do not contradict the comparison sorting lower bound.^[ref]