This is about the *only* proof I can come up with for **Sperner's theorem** (which see). It's an interesting question whether other, structurally different, proofs exist. In particular, a truly probabilistic proof (not just rephrasing the counting argument below in terms of probability) would be desirable (to me, at least).

### Proof.

Suppose we have some **antichain** F on {1,...,n}. Consider all ordered **permutations** a_{1},a_{n}∈{1,...,n} (that is, all *orderings* of the numbers 1,...,n). There are n! such orderings. Now, the sequence of sets

{acontains at most one element of F (otherwise F is no antichain)._{1}} ⊆ {a_{1},a_{2}} ⊆ ... ⊆ {a_{1},a_{2},...,a_{n}}

Suppose f∈F has k elements. Then there are exactly k!(n-k)! orderings that intersect f. So, if c_{k} is the number of elements of F with k elements, by intersections of orderings we have that

∑Now, we're interested in just ∑c_{k=0}^{n}c_{k}⋅k!(n-k)! ≤ n!

_{k}= |F| (the number of elements of F). But for all k we have that

k!(n-k)! ≥ ⌊n/2⌋!⌈n/2⌉! = m(n)(this is just the claim that the largest binomial coefficient is the central one). So,

|F|⋅m(n) = ∑or, in other words, |F| is not larger than the central binomial coefficient._{k=0}^{n}c_{k}⋅m(n) ≤ ∑_{k=0}^{n}c_{k}⋅k!(n-k)! ≤ n!,

**QED.**