Condition on Lower Coefficient for Binomial Coefficient to be Maximum

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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.

$\blacksquare$


Sources