summation of products of combinatorials For any natural number $N$ and $0\le n\le N$ define
$$ 
f(n) = f(n,N) = \frac{1}{(N+1)!} \sum_{\substack{{S\subset \{1,\ldots,N\}} \\ {|S|=n}}} \prod_{s\in S} s. 
$$
(The empty product is interpreted as $1$.)  It is easy to see that 
$$
\sum_{n=0}^{N} f(n) = 1, 
$$ 
so that $f$ may be thought of as a probability distribution. 
As $N$ gets large, what is the maximum value of $f(n)$?  What is the value $n_{max}$ where this peak value is attained, and 
what is the distribution of $f$, especially around $n_{max}$?   
 A: The unnormalized sums are unsigned Stirling numbers of the first kind, the coefficient of $x^{N-n}$ in $(x+1)(x+2)\cdots(x+N).$
For $p=1/2$, this is the central Stirling number of the first kind $3,35,735,22449, ..., S_1(2n-1,n), ...$. See A129505 which mentions
$$S_1(2n-1,n) \sim \frac{1+2c}{8c\sqrt{-\pi(1+c)}} \bigg(\frac{-8c^2}{e(1+2c)}\bigg)^n n^{n-3/2}$$
where $c=\operatorname{LambertW}(-1,\frac{-1}{2\sqrt{e}}) \sim -1.75643$ 
in a note updated about a month ago. 

When you think about this as a probability distribution, it is a sum of independent Bernoulli random variables with probabilities $1/2, 2/3, ... N/(N+1)$. It's easy to read off the mean $N+2-H_{N+1} \approx N-\log N + c_1$ and variance $1/4 + 2/9 + ... + N/(N+1)^2 \approx \log N + c_2$. Lindeberg's condition is satisfied so the distribution is asymptotically normal. 
Each binomial factor is unimodal, so the product is unimodal, so the asymptotic normality with standard deviation asymptotic to $\sqrt{\log N}$ gives asymptotic lower bounds for the maximum probability asymptotic to $\frac{1}{\sqrt{2 \pi \log N}},$ which agrees with The Masked Avenger's heuristics in the comments. I think there should be asymptotic upper bounds of the same form. 
A: Although the problem already has an accepted answer, this may be a useful connection.  As Douglas Zare notes the problem asks for an understanding of the coefficients of $(x+1)(x+2)\cdots (x+N)$.  Now the $x^k$-th coefficient of the polynomial $x(x+1)\cdots(x+N)$ counts the number of permutations in $S_{N+1}$ with $k$ cycles.  The distribution of the number of cycles in a random permutation has been extensively studied, and is well known to be Gaussian with mean about $\log N$ and variance about $\log N$ (by work of Goncharov 
from the 1940s).  Thus, in the notation of the problem $f(n)$ is the probability that a 
random element of $S_{N+1}$ has $N+1-n$ cycles.  
