Given positive integers $n$ and $c$, with $n>c$, is there a good way to estimate the number of ways to partition the set $\{1,\dots,n\}$ into ordered subsets, with each subset having at most $c$ elements? By "ordered subsets", I mean that the ordering *within* each subset matters, but we don't care about the order in which the subsets are presented to us, e.g. the partitions $\{1,2,3\},\{4,5,6\}$ and $\{3,2,1\},\{4,5,6\}$ are distinct, but $\{1,2,3\},\{4,5,6\}$ and $\{4,5,6\},\{1,2,3\}$ are the same. How about if we only consider partitions into subsets whose sizes are all between $c/2$ and $c$?


As an aside, in my actual problem, I am interested in the asymptotic behavior as $n\to\infty$ with $c=a\sqrt{n}$, with $a$ a constant.



(Approximate) Number of ways to partition $n$ elements into ordered subsets of bounded size
Let $S = \{(a\_1, \ldots , a\_c) \ : \ \sum\_i i a\_i = n\}$, and for $a \in S$, let $f(a) = 1/\prod\_i a\_i !$. Then the exact value you want is $n! \sum\_{a \in S} f(a)$ [the term $n! f(a)$ counts the number of such decompositions with $a\_i$ sets of size $i$].


We can bound this by finding some $f(a) \leq \Delta$, which would give


$$\Delta \leq N / n! \leq |S| \Delta = p\_c (n) \Delta$$


where $N$ is the number you want, and $p\_c (n)$ is the number of partitions of $n$ into parts of size at most $c$. (Then use Stirling's formula and some useful upper bound on like perhaps $p\_c (n) \leq p(n) \sim \frac{1}{4n \sqrt{3}} \exp[\pi \sqrt{2n/3}].$)


Is that good enough? If not, you could get more mileage out of these bounds.


**Added:**


Or in the case where each set is to have size between $c/2$ and $c$, define the same $S$ as before (except insisting $a\_i =0$ for $i<c/2$). Then for $a \in S$, we have $n/c \leq \sum a\_i \leq 2n/c$. Thus


$$N \leq \sum\_{a \in S} n! / f(a) \leq (n-2n/c)! \sum\_{a \in S} n! / [f(a) (n-\sum\_i a\_i)!] \leq (n-2n/c)! (2+c/2)^n.$$


And we can then combine this with $N \geq n! / (n/c)!$ to get a moderately decent approximation.


Is that close enough?


**Second edit:**


Or in the above setting, we can use $|S| \leq (n/c) p\_{c} (2n/c)$ and then follow as in the first idea.


**Third edit:**


Let $g(n,c)$ denote the number of decompositions into sets of size at most $c$. Then we have the exponential generating function:


$$G\_c (z) = \sum\_{n} \frac{g(n,c)}{n!} z^n = \prod\_{k=1}^{c} \sum\_{j=0} ^{\infty} \frac{z^{jk}}{j!} = \prod\_{k=1}^{c} e^{z^k} = e^{z (z^c -1)/(z-1)}.$$


Thus, we are trying to estimate $G\_c ^{(n)} (0)$. (Just a thought)


