Given a set of $2n$ disks with $n \geq 10$. For every radius $j=1,...,n$ there is one pair of disks. One of the disks is transparent, and the other one is not. One of the disks is chosen for every pair. The disks are arbitrarily placed on a stake. In how many ways can you choose and stack the disks such that all the edges of the $n$ disks are visible from above?
I thought there are $2^n$ ways to choose the disks (you can either choose the transparent or the opaque disk for all of the $n$ pairs). If you choose only opaque disks, then you there is only $1$ option to place the disks. If you choose only transparent disks, then it doesn't matter what order they are stacked, so you can stack them in $n!$ ways. But I am not sure how I can find a general expression for the other cases.
Let $f(n)$ be the total number of arrangements with $n$ discs. For $n=1$ there are two arrangements - a single transparent disc or a single opaque disc.
Suppose we are given a stack with $n$ discs, and are adding a larger disc to it somewhere. If the large disc we are adding is opaque, then it can only go on the bottom. If it is transparent, then we can add it at any of the $n+1$ places on/between/under the discs of the stack. This means that $f(n+1) = (n+2)*f(n)$.
So
$f(1)=2$
$f(2)=2*3=6$
$f(3)=2*3*4=24$
$f(4)=2*3*4*5=120$
...
This recurrence relation pretty obviously gives you $f(n) = (n+1)!$.