Question: Let $R$ be a noetherian ring and $M$ be a finitely generated $R$ - Module.

Prove that there exists $n\in \mathbb N$ such that there is a finitely generated $R$ sub-module $N\subset R^n$ such that 

$$R^n / N \cong M$$

Any ideas ? Maybe use the isomorphism theorem
?

Answer: 
$$\frac{(2n)!-2(n!)^2}{(n!)^2 n}
=
\frac{1}{n} \left( \frac{(2n)!}{(n!)^2} - 2 \right)
=
\frac{1}{n} \left( \binom{2 n}{n} - 2 \right).$$

So one has to prove that if $n$ is prime, then 
$$
\binom{2 n}{n} \equiv 2 \pmod{n}.
$$
This is a very special case of [Lucas's Theorem](https://en.wikipedia.org/wiki/Lucas's_theorem). In this particular instance, just [compute](https://en.wikipedia.org/wiki/Frobenius_endomorphism) in $\mathbb{Z}/ n\mathbb{Z} [x]$
$$
\sum\_{i=0}^{2n} \binom{2n}{i} x^{i} = (1 + x)^{2n} = (1 + x^{n})^{2} = 1 + 2 x^{n} + x^{2 n}
$$
and compare the coefficients of $x^{n}$.
This simplifies to $\big(\binom{2n}n-2\big)/n$. 

Now $\binom{2n}{n}$ is the number of ways of choosing $n$ nuumbers from $1,...,2n$. If we choose $r$ from $1,...,n$ we have to choose $n-r$ from $n+1,...,2n$, and so this equals $\sum\_{r=0}^n\binom{n}{r}\binom{n}{n-r}$. 

So $\binom{2n}n-2=\sum\_{r=1}^{n-1}\binom{n}{r}\binom{n}{n-r}$ (the two terms we've missed out are both $1$), and since $n$ is prime it is easy to check that all terms on the RHS are divisible by $n$.