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 $6n^3$?
 A: You have, as $n \to \infty$,
$$
\frac{n^4}{\binom{4n}{4}}=\frac{n^4 4! (4n-4)!}{(4n)!}=\frac{\color{red}{24n^4}}{\color{red}{4n}(4n-1)(4n-2)(4n-3)}=\frac{\color{red}{6n^3}}{(4n-1)(4n-2)(4n-3)}.
$$
A: You have a little mistake with the binomial coefficient:
$$\frac{n^4}{\binom{4n}4}=\frac{4!n^4(4n-4)!}{(4n)!}=\frac{24n^4}{4n(4n-1)(4n-2)(4n-3)}\xrightarrow[n\to\infty]{}\frac{24}{256}=\frac3{32}$$
A: Hint:
$$
\begin{align}
\frac{n^4}{\binom{4n}{4}}
&=\frac{n^4}{\frac{4n(4n-1)(4n-2)(4n-3)}{4!}}\\
&=\frac{n^4}{\frac{4^4n^4\left(1-\frac1{4n}\right)\left(1-\frac2{4n}\right)\left(1-\frac3{4n}\right)}{4!}}\\
&=\frac{4!}{4^4}\frac1{\left(1-\frac1{4n}\right)\left(1-\frac2{4n}\right)\left(1-\frac3{4n}\right)}
\end{align}
$$
A: 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, $$\lim_{n \to \infty} a_n(m) = \prod_{k=1}^m \lim_{n \to \infty} \frac{k}{m+(k-m)/n} = \prod_{k=1}^m \frac{k}{m} = \frac{m!}{m^m}.$$
