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:
 * $k_{\max} = \left\lfloor{\dfrac n 2}\right\rfloor$

and also:
 * $k_{\max} = \left\lceil{\dfrac n 2}\right\rceil$

Proof
From Condition for Increasing Binomial Coefficients:
 * $\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1}\iff k \le \dfrac n 2$

Let $n$ be an even integer, such that:
 * $n = 2 r$

Then:
 * $r = \left\lfloor{\dfrac n 2}\right\rfloor = \left\lceil{\dfrac n 2}\right\rceil$.

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

From Symmetry Rule for Binomial Coefficients:
 * $\forall k < r: \dbinom n r > \dbinom n {n - k}$

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:
 * $k_{\max} = \left\lfloor{\dfrac n 2}\right\rfloor = \left\lceil{\dfrac n 2}\right\rceil$

Let $n$ be an odd integer, such that:
 * $n = 2 r + 1$

Then:
 * $r = \left\lfloor{\dfrac n 2}\right\rfloor$

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:
 * $r = \left\lfloor{\dfrac n 2}\right\rfloor$

and:
 * $r + 1 = \left\lceil{\dfrac n 2}\right\rceil$

Hence the result.