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 a little mistake with the binomial coefficient:
$$\frac{n^4}{\binom{4n}4}=\frac{4!n^4(4n-4)!}{(4n)!}=\frac{24n^4}{4n(4n-1)(4n-2)(4n-3)}\xrightarrow[n\to\infty]{}\frac{24}{256}=\frac3{32}$$
