Condition for Equality of Adjacent Binomial Coefficients

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Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $\dbinom n k$ denote a binomial coefficient for $k \in \Z$.


Then:

$\dbinom n k = \dbinom n {k + 1}$

if and only if:

$n$ is an odd integer
$k = \dfrac {n - 1} 2$


Proof

Sufficient Condition

Let $n$ be odd and $k = \dfrac {n - 1} 2$.

Let $n = 2 m + 1$ for some $m \in \Z_{\ge 0}$.


We have:

\(\ds k\) \(=\) \(\ds \dfrac {n - 1} 2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {2 m + 1} - 1} 2\)
\(\ds \) \(=\) \(\ds \dfrac {2 m} 2\)
\(\ds \) \(=\) \(\ds m\)


Hence:

\(\ds \dbinom n k\) \(=\) \(\ds \dbinom {2 m + 1} m\)
\(\ds \) \(=\) \(\ds \dbinom {2 m + 1} {\paren {2 m + 1} - m}\) Symmetry Rule for Binomial Coefficients
\(\ds \) \(=\) \(\ds \dbinom {2 m + 1} {m + 1}\)
\(\ds \) \(=\) \(\ds \dbinom n {k + 1}\)

$\Box$


Necessary Condition

Let $n$ and $k$ be such that:

$\dbinom n k = \dbinom n {k + 1}$

From Condition for Increasing Binomial Coefficients, it is not the case that:

$0 \le k < \dfrac {n - 1} 2$

as under such a condition we would have:

$\dbinom n k = \dbinom n {k + 1}$

So:

$k \ge \dfrac {n - 1} 2$

By the Symmetry Rule for Binomial Coefficients, we also have that:

$\dfrac {n - 1} 2 < k \le n \iff \dbinom n k > \dbinom n {k + 1}$

and so for $\dbinom n k = \dbinom n {k + 1}$ it is not the case that $\dfrac {n - 1} 2 < k \le n$.

So:

$k \le \dfrac {n - 1} 2$

Hence that means:

$k = \dfrac {n - 1} 2$

which means:

$n = 2 k + 1$

and so $n$ is odd.

$\blacksquare$


Sources

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