A neater solution to an interesting question part 2 This is a continuation from another question given here:
A neater solution to an interesting question
I reproduce the problem statement here:
Just the other day, a discussion with a friend prompted the following question: There is a beautiful lady who works at the cashier. A person can only see all the way to the front of the queue if they are taller than each person in front of them, and every individual is either strictly taller or strictly shorter than any one else in the queue. 
Part 1 of the question was to determine the expected number of people who can see the cashier. The solution is given by linearity of expectation and noting that the probability that the kth person can see to the front (by virtue of being the tallest among k) is 1/k.
Part 2 of the question is as follows: What is the probability that m out of n people can see to the front?
The solutions I have arrived at so far are as follows:
$\Bbb P\ (\ number \ able\ to \ see =m) = 
\frac{1}{n} 
\sum_{k_2=m-1}^{n-1}\frac{1}{k_2}( 
\sum_{k_3=m-2}^{k_2-1}\frac{1}{k_3}(...
(\sum_{k_m=1}^{k_{m-1}-1}\frac{1}{k_m})...)) $
The subscript of $k_i$ corresponds to the tallest person in front of the $k_{i-1}$th tallest person. The value of $k_i$ denotes the number of people in front of the $k_{i-1}$th tallest person, and hence $\frac{1}{k_i}$ denotes the probability that that $k_i$th person is in any particular position.

Given user49640's suggestion, I include a recursive expression of the above solution for $\Bbb P_n$. n denotes the number of people in the queue, and m denotes the number of people who can see to the front.
$\Bbb P_n\ (no. =m) = \frac{1}{n} \sum_{k=m-1}^{n-1}\Bbb P_k\ (no. = m-1) $

My question is: Is this correct? If so, is there a neater or better solution?
 A: The recursive expression you give is correct. The probability that $k$ out of $n$ people can see the cashier is given by $\frac{1}{n!} \left[{n \atop k}\right]$, where $\left[{n \atop k}\right]$ is a Stirling number of the first kind (unsigned). 
Usually, Stirling numbers of the first kind are used to count permutations with $k$ cycles. This is equivalent to your formulation because each cycle decomposition can be uniquely written with the highest number in each cycle put first, and then the cycles sorted by their first number in ascending order. (So we could write the cycle decomposition $(1\,3\,4)\,(2\,5)\,(6)$ as $(4\,1\,3)\,(5\,2)\,(6)$.) 
Flattening this out gives a permutation of $n$ elements in which exactly $k$ are the largest seen so far: in the example, we get $4 1 3 5 2 6$, in which persons $4$, $5$, and $6$ can see the cashier at the front. Conversely, given a permutation, we can take the people who can see the cashier, and interpret them as beginning a cycle, obtaining a cycle decomposition written in the above form; so this gives a bijection.
Wikipedia lists the recurrence $\left[{n+1 \atop k}\right] = n\left[{n \atop k}\right] + \left[{n \atop k-1}\right]$ for these; this gives $$\mathbb P_{n+1}(k) = \frac{n}{n+1} \mathbb P_n(k) + \frac1{n+1} \mathbb P_n(k-1)$$ for the probability. This can probably be derived from your recurrence, and also explained as follows: when there are $n+1$ people, with probability $\frac1{n+1}$ the very tallest person is at the back of the line (so if we wanted $k$ people to see the cashier, we have $k-1$ left to find); otherwise, the person at the back of the line can be ignored.
There are plenty of other identities on Wikipedia, but not a nice closed form or anything.
