<s>[INST] Bounds for $\binom{n}{cn}$ with $0 < c < 1$. Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$?  I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn}  \leq \left(\frac{e}{c}\right)^{cn}$.
 [/INST] Stirling's Asymptotic Expansion, derived here, is
$$
n!=\sqrt{2\pi n}\,n^ne^{-n}\left(1+\frac1{12n}+\frac1{288n^2}-\frac{139}{51840n^3}-\frac{571}{2488320n^4}+O\left(\frac1{n^5}\right)\right)
$$
From which we get
$$
\begin{align}
&\frac{n!}{(cn)!((1-c)n)!}\\[6pt]
&=\frac{\left(c^c(1-c)^{1-c}\right)^{-n}}{\sqrt{2\pi c(1-c)n}}\small\left(1-\frac{1-c+c^2}{12c(1-c)}\frac1n+\frac{1-2c+3c^2-2c^3+c^4}{288c^2(1-c)^2}\frac1{n^2}+O\left(\frac1{n^3}\right)\right)
\end{align}
$$

Absolute Bounds
For $n\ge1$,
$$
\sqrt{2\pi n}\,n^ne^{-n}\left(1+\frac1{12n}\right)\le n!\le\sqrt{2\pi n}\,n^ne^{-n}\left(1+\frac1{12n}+\frac1{288n^2}\right)
$$
which gives
$$
\sqrt{2\pi n}\,n^ne^{-n}\le n!\le\frac{313}{288}\sqrt{2\pi n}\,n^ne^{-n}
$$
From which we get, for $1\le cn\le n-1$,
$$
\left(\frac{288}{313}\right)^2\frac{\left(c^c(1-c)^{1-c}\right)^{-n}}{\sqrt{2\pi c(1-c)n}}
\le\binom{n}{cn}
\le\frac{313}{288}\frac{\left(c^c(1-c)^{1-c}\right)^{-n}}{\sqrt{2\pi c(1-c)n}}
$$
Note that
$$
\frac{e}{\sqrt{2\pi}}\doteq1.0844\lt1.0868\doteq\frac{313}{288}
$$
so this is close to Hagen von Eitzen's answer.

The bound mentioned in Hagen's answer is based on the fact that the ratio
$$
\frac{\Gamma(n)}{\sqrt{2\pi n}\,n^ne^{-n}}
$$
is decreasing in $n$. Thus, the greatest ratio is for $n=1$ and therefore,
$$
\sqrt{2\pi n}\,n^ne^{-n}\le n!\le\frac{e}{\sqrt{2\pi}}\sqrt{2\pi n}\,n^ne^{-n}
$$
</s>