# Condition on Lower Coefficient for Binomial Coefficient to be Maximum

## Theorem

Let $n, k \in \Z_{\ge 0}$ be positive integers.

Let $\dbinom n k$ denote the binomial coefficient of $n$ choose $k$.

Then for a given value of $n$, the value of $k$ for which $\dbinom n k$ is a maximum is:

$k_\max = \floor {\dfrac n 2}$

and also:

$k_\max = \ceiling {\dfrac n 2}$

## Proof

$\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1} \iff k \le \dfrac n 2$

Let $n$ be an even integer, such that:

$n = 2 r$

Then:

$r = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$

and:

$\forall k < r: \dbinom n r > \dbinom n k$
$\forall k < r: \dbinom n r > \dbinom n {n - k}$

and so:

$\forall k > r: \dbinom n r > \dbinom n k$

We also have:

$\dbinom n r = \dbinom n {n - r} = \dbinom n {n / 2}$

Thus for all $k \ne r$:

$\dbinom n k < \dbinom n {n / 2}$

and so:

$k_\max = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$

$\Box$

Let $n$ be an odd integer, such that:

$n = 2 r + 1$

Then:

$r = \floor {\dfrac n 2}$

As $r < n$:

$\forall k < r: \dbinom n r > \dbinom n k$

When $k = r$ we also have:

$\dbinom n r = \dbinom n {n - r} = \dbinom n {r + 1}$

Then from Symmetry Rule for Binomial Coefficients:

$\forall k < r: \dbinom n r > \dbinom n {n - k}$

and so:

$\forall k > r + 1: \dbinom n r > \dbinom n k$

We have that:

$r = \floor {\dfrac n 2}$

and:

$r + 1 = \ceiling {\dfrac n 2}$

Hence the result.

$\blacksquare$