LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31

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The binomial coefficients are symmetric and increase to the middle level before decreasing

Statement

For every $n$ and $0 \leq k \leq n$ ,

$(\binom{n}{k}) = (\binom{n}{n - k}) .$

For $0 \leq k < n$ ,

$(\binom{n}{k + 1}) \geq (\binom{n}{k}) ⟺ k + 1 \leq n - k .$

Consequently the binomial coefficients increase up to the middle rank and decrease after it. Their maximum is attained only at $k = n / 2$ when $n$ is even, and at the two ranks $k = (n - 1) / 2$ and $k = (n + 1) / 2$ when $n$ is odd.

Facts & Assumptions

Given: Natural numbers $n$ and $k$ with $0 \leq k \leq n$ .

[F1]

$(\binom{n}{k})$ is the number of $k$ -element subsets of an $n$ -element set (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L1]

The binomial closed formula gives symmetry and, for $k < n$ , the identity $(\binom{n}{k}) (n - k) = (\binom{n}{k + 1}) (k + 1)$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

[L2]

The product rule licenses the usual double count of a set together with a chosen element outside or inside it (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

Proof

technique · direct

1.1

The symmetry $(\binom{n}{k}) = (\binom{n}{n - k})$ is the symmetry clause of [L1].

1.2

For $k < n$ , count pairs $(S, x)$ with $∣ S ∣ = k$ and $x \notin S$ by first choosing $S$ , or by first choosing the $(k + 1)$ -set $S \cup {x}$ and then the deleted element. This gives $(\binom{n}{k}) (n - k) = (\binom{n}{k + 1}) (k + 1)$ , in agreement with [L1].

F1L1L2

2.1

Since both $n - k$ and $k + 1$ are positive, step 1.2 shows that $(\binom{n}{k + 1}) \geq (\binom{n}{k})$ exactly when $n - k \geq k + 1$ , with equality exactly when $n - k = k + 1$ .

step 1.2algebra

3.1

Reading step 2.1 as $k$ increases gives strict increase before the middle, equality between the two middle ranks only when $n$ is odd, and strict decrease afterward; symmetry from step 1.1 identifies the stated maximizing ranks.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

M. Keller and W. T. Trotter, Applied Combinatorics, §6.2 (standard reference, not scraped)