Given an set of unique integers of length N. What are number of different ways you can rearrange the array so that, you can only see X numbers of integers from the left and Y numbers of integers from the right, with the restriction being that the biggr integer in the set blocks the smaller integers. So if you had say [10,40,60,20], you would see 3 from the left and 2 from the right.
Example:
Given: N = 3, X = 2, Y = 2 Output: 2
Given: N = 6, X = 1, Y = 2 Output: 24
You get a tractable recurrence if, instead of considering where the largest number is, you look at the smallest number.
The indexing will seem strange here, but bear with me. Let $f(n,l,r)$ denote the number of permutations of $\{0,\ldots,n\}$ where $l+1$ elements are visible from the left and $r+1$ are visible from the right. Clearly $f(n,l,r)=0$ if either $l$ or $r$ is negative, since at least one element (the largest) is visible on each side. We have $f(0,l,r)=[l=r=0]$. Also, the only way the smallest element is visible is if it is at one end, so we get the recurrence $$f(n,l,r)=f(n-1,l-1,r)+f(n-1,l,r-1)+(n-1)f(n-1,l,r)$$ for $n>1$.
It's not hard to show inductively that $$f(n,l,r)=\binom{l+r}{r}\,|S_n^{(l+r)}|$$ where $S_n^{(k)}$ denotes the Stirling number of the first kind. (There's probably a simple direct argument but I haven't had time to think about that.)
In particular, the two given examples are $\binom21\,|S_2^{(2)}|=2$ and $\binom10\,|S_5^{(1)}|=24$.