# How fast does $\binom{n}{k}$ grow when $k \le n/2$?

How fast does $\binom{n}{k}$, $n$ fixed, grow when $k \le n/2$?

Especially, I'm interested in the growth of the "inverse" of binomial coefficient $B_n(x) := \min \{k:\binom{n}{k} \ge x\}$.

EDIT: This was answered in a previous post, according to which $B_n(x) = O(\log x)$.

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hardmath: My application of the analysis of an algorithm whose inner loop is run $B_n(x)$ times. So, as you've said, the ratio or difference between two consecutive terms is a good measure, but it doesn't fit my application well. By the way, $x$ is much smaller than a typical value of $\binom{n}{k}$, so what I'd like to know is the growth of $\binom{n}{k}$ when $k$ is small. – Pteromys Jan 30 '12 at 14:56
On your initial question, $$\dfrac{\binom{n}{k}}{\binom{n}{k-1}} = \dfrac{n-k}{k} = \dfrac{n}{k}-1$$ which grows faster when $k$ is small, and is close to $1$ when $k$ is close to $n/2$. – Henry Jan 30 '12 at 15:05