Here's what I think you might be thinking of: Stirling numbers of the first kind count permutations according to their number of cycles. But the mean and variance of the number of cycles of permutations on $n$ elements are well-known. In particular, given a random variable $X_n$ which represents that number of cycles, $E(X_n) = 1 + 1/2 + \cdots + 1/n$ and $Var(X_n) = E(X_n) - (1 + 1/2^2 + \cdots + 1/n^2)$. In fact $X_n$ can be written as the sum of n independent Bernoulli random variables, where the kth such variable has mean 1/k. This decomposition then lets you invoke the Central Limit Theorem and say that the distribution is, in some limiting sense, normal.
So say we want to use this to estimate, for example, $S(9,3)$. The average number of cycles of a permutation on 9 elements is $1 + 1/2 + \cdots + 1/9 \approx 2.828968$; call this $\mu$. The variance of the number of cycles is $\mu - (1 + 1/2^2 + \cdots + 1/9^2) \approx 1.289201$; call this $\sigma^2$. Then by this normal approximation, the probability that a random permutation on 9 elements has 3 cycles is
$\Phi \left( {3.5 - \mu \over \sigma} \right) - \Phi \left( {2.5 - \mu \over \sigma} \right) $
where $\Phi$ is the standard normal CDF. This is $0.336726$; so we estimate S(9,3) is 9! times this, or 122191. The actual value is 118124, so this isn't a bad guess.
However, using the same method to estimate S(9,8) gives about 7 when the true value is 36; as with any normal approximation, it's better towards the center of the distribution than in the tails.