Editing Binomial coefficient (section)

=== Both ''n'' and ''k'' large ===
[[Stirling's approximation]] yields the following approximation, valid when <math>n-k,k</math> both tend to infinity:
<math display="block">{n \choose k} \sim
\sqrt{n\over 2\pi k (n-k)} \cdot
{n^n \over k^k (n-k)^{n-k}}
</math>
Because the inequality forms of Stirling's formula also bound the factorials, slight variants on the above asymptotic approximation give exact bounds.
In particular, when <math>n</math> is sufficiently large, one has
<math display="block"> {2n \choose n} \sim \frac{2^{2n}}{\sqrt{n\pi }}</math> and <math>\sqrt{n}{2n \choose n} \ge 2^{2n-1}</math>. More generally, for {{math|''m'' ≥ 2}} and {{math|''n'' ≥ 1}} (again, by applying Stirling's formula to the factorials in the binomial coefficient),
<math display="block">\sqrt{n}{mn \choose n} \ge \frac{m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}}.</math>

If ''n'' is large and ''k'' is linear in ''n'', various precise asymptotic estimates exist for the binomial coefficient <math display="inline"> \binom{n}{k}</math>. For example, if <math>| n/2 - k | = o(n^{2/3})</math> then
<math display="block"> \binom{n}{k} \sim \binom{n}{\frac{n}{2}} e^{-d^2/(2n)} \sim \frac{2^n}{\sqrt{\frac{1}{2}n \pi }} e^{-d^2/(2n)}</math>
where ''d'' = ''n'' − 2''k''.<ref>{{cite book|title=Asymptopia|last1=Spencer|first1=Joel|last2=Florescu|first2=Laura|date=2014| publisher=[[American Mathematical Society|AMS]]|isbn=978-1-4704-0904-3|series=Student mathematical library|volume=71|page=66| oclc=865574788|author1-link=Joel Spencer}}</ref>