<s>[INST] Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this is the part I don't understand
$\rightarrow \frac{6}{32}$
How did the numerator simplify to $6n^3$?
 [/INST] Hint:
$$
\begin{align}
\frac{n^4}{\binom{4n}{4}}
&=\frac{n^4}{\frac{4n(4n-1)(4n-2)(4n-3)}{4!}}\\
&=\frac{n^4}{\frac{4^4n^4\left(1-\frac1{4n}\right)\left(1-\frac2{4n}\right)\left(1-\frac3{4n}\right)}{4!}}\\
&=\frac{4!}{4^4}\frac1{\left(1-\frac1{4n}\right)\left(1-\frac2{4n}\right)\left(1-\frac3{4n}\right)}
\end{align}
$$
</s>