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 $6n^3$?

It is worth generalizing the given limit:  for positive integers $m$,
$$\begin{align*} a_n(m) &= \frac{n^m}{\binom{mn}{m}} \\ &= \frac{n^m m! \, (m(n-1))!}{(mn)!} \\ &= \frac{n^m m!}{(mn)(mn-1)\cdots(m(n-1)+1)} \\ &= \prod_{k=1}^m \frac{nk}{m(n-1) + k} \\ &= \prod_{k=1}^m \frac{k}{m + (k-m)/n}\end{align*}.$$  Consequently, $$\lim_{n \to \infty} a_n(m) = \prod_{k=1}^m \lim_{n \to \infty} \frac{k}{m+(k-m)/n} = \prod_{k=1}^m \frac{k}{m} = \frac{m!}{m^m}.$$
