binomial coefficient aproximation

Question

I found this aproximation

$$ \binom{n}{k}\sim \frac{n^k}{k!} $$

But there was a note that said this holds just when $n>>k$, why is this? I need to consider this binomial coefficient when $k$ is near of $n/2$ and I would like use this or any similar expression to avoid some factorials.

yes, that's exactly what I want to use, but I just want to know if I can reduce the factorial expressions in order to have less Stirling's — Luis GC, Mar 19 '18 at 14:54

score 2 · Accepted Answer · answered Mar 19 '18 at 14:55

You can write the coefficient as $\frac {n(n-1)(n-2)\ldots (n-k+1)}{k!}$ When $n \gg k$ the numbers you subtract from $n$ in the numerator may be too small to matter. If you ignore them you get the approximation $\frac {n^k}{k!}$ When $k$ is near $\frac n2$ this is not a good approximation. Stirling's approximation is quite accurate, leading to the approximation of the central binomial coefficient $${2n \choose n} \approx \frac {4^n}{\sqrt {\pi n}}$$

score 0 · Answer 2 · answered Mar 19 '18 at 14:57

0

Because when $k$ is not a lot smaller than $n$, it doesn't work. We see this, for instance, with $k = n$ where $\binom nk = 1$ but $\frac{n^k}{n!}\gg 1$.

answered Mar 19 '18 at 14:57

Arthur

199,419

binomial coefficient aproximation

2 Answers2