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 $6n^3$?

You have, as $n \to \infty$,
$$
\frac{n^4}{\binom{4n}{4}}=\frac{n^4 4! (4n-4)!}{(4n)!}=\frac{\color{red}{24n^4}}{\color{red}{4n}(4n-1)(4n-2)(4n-3)}=\frac{\color{red}{6n^3}}{(4n-1)(4n-2)(4n-3)}.
$$
