TITLE: permutations of a multiset having symbols with fixed multiplicity
QUESTION [8 upvotes]: Let $N$ be a multiset of $n$ distinct objects having the same multiplicity $k$. For instance,
$N=\{a,\,a,\,b,\,b\}$
where $n=2$ and $k=2$.
I was looking for the problem of counting the number of all the permutations of all the non-empty subsets of $N$.
For the N in the above example, there are 18 permutations of the subsets of N
$$
a,b,\;\;
aa,ab,ba,bb,\;\;
aab,aba,abb,baa,bab,bba,\;\;
aabb,abab,abba,baab,baba,bbaa$$
Till now I've ended up just with an upper bound of that number, by using the following idea.
Compute all of the permutations of $N$ of length $nk$ (ignore subsets) with the basic formula
$$
perm(N;nk)=\frac{(nk)!}{\underbrace{(k!)(k!)...(k!)}_{n \; times}}
$$
Then, just take all the $nk$-prefixes of all the objects in $perm(N;nk)$. That gives
$$
nk\frac{(nk)!}{(k!)^{n}}
$$
However since this method counts twice a prefix shared by two $nk$-permutations (e.g. $ab$), I end up with a coarse upper bound. For $N$ in the example, $24$ ($>18$).
I was wondering if I can find a closed formula that computes exactly the desired number ?

REPLY [2 votes]: I calculated the following table of values with GAP:
$$\tiny \begin{array}{r|rrrrrr}
    & n=1 & 2 & 3 & 4 & 5 & 6 \\
\hline
k=1 & 1 & 4 & 15 & 64 & 325 & 1956 \\
2 & 2 & 18 & 270 & 7364 & 326010 & 21295782 \\
3 & 3 & 68 & 5247 & 1107696 & 492911195 & 396643610628 \\
4 & 4 & 250 & 110250 & 191448940 & 904434761800 & 9459612561834054 \\
5 & 5 & 922 & 2435199 & 35899051100 & 1856296498826905 & 260080189398141873510 \\
6 & 6 & 3430 & 55621566 & 7101534312684 & 4098746255797339510 & 7843008902239185171370146  \\
7 & 7 & 12868 & 1301226495 & 1458965717496880 & 9527671240728144876535 & 252375551579644513763596356612 \\
8 & 8 & 48618 & 30992872482 & 308290573348183628 & 23008745547185924987336820 & 8521137082687484698075500756496758 \\
9 & 9 & 184754 & 748574130015 & 66577182435768923244 & 57224682697609859297931802325 & 298568465318687485121812233253672384734 \\
10 & 10 & 705430 & 18283414868862 & 14629025943480502591444 & 145692364455943850433130828021810 & 10773961436670577098657278942204092992628650 \\
\end{array}$$
Some of these numbers don't appear in Sloane's OEIS (or even Google [yet]), and many of those that appear belong to sequences without references (such as A144660 or A144661).  So it's unlikely there's a known formula, but we might be able to tackle some simple cases.

When $n=2$ the number is $\binom{2k+2}{k+1}-2$, which is a special case of Dennis Gulko's answer, simplified using the binomial identity $$\sum_{s=0}^k \sum_{t=0}^k \binom{s+t}{s} = \binom{2k+2}{k+1}-1$$ (see Sloane's A030662).
Sloane's A003011 gives formulae for the $k=2$ case, e.g. a recurrence relation.