I am familiar with:
$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$
Now, i have an excercise in calculus which gets to the point that we should calculate:
$$ \frac{\binom{3n+1}{n+1}}{\binom{3n}{n}} $$
The answer says that:
$$ \frac{\binom{3n+1}{n+1}}{\binom{3n}{n}} = \frac{(3n+3)!n!(2n)!}{(3n)!(2n+2)!(n+1)!} $$
I dont understand how they got to thise from the formula i showed, how, for example, they got to $(3n+3)$, $(2n+2)$?
Looking at the result, I think the original questions was meant to show,
$$ \frac{ \left( 3(n+1) \atop n+1 \right) } { \left( 3n \atop n \right) } = \frac{(3n+3)!}{(n+1)!(2n+2)!} \cdot \frac{ n!(2n)! }{(3n)!}. $$
which is simply applying the formula.
The expression you have simplifies to $$\frac{3n+1}{n+1}.$$