How many ways to place $n$ distinguishable balls into $m$ distinguishable bins of size $s$?

Question

Let there be $n$ distinguishable balls and $m$ distinguishable bins, each bin of size $s$, that is, we cannot place more than $s$ balls into it. How many possibilities are there to place the balls into the bins?

Answer 1

This can be done using species and generating functions. If some bins are allowed to remain empty the species is

$$\mathfrak{S}_{=m}(\mathfrak{P}_{\le s}(\mathcal{Z})).$$

This gives the generating function $$P(z) = \left(\sum_{k=0}^s \frac{z^k}{k!}\right)^m$$

and the closed formula $$p_{n,m,s} = n! [z^n] \left(\sum_{k=0}^s \frac{z^k}{k!}\right)^m.$$

If no bins are allowed to remain empty the species is

$$\mathfrak{S}_{=m}(\mathfrak{P}_{1\le\cdot\le s}(\mathcal{Z})).$$

This gives the generating function $$Q(z) = \left(\sum_{k=1}^s \frac{z^k}{k!}\right)^m$$

and the closed formula $$q_{n,m,s} = n! [z^n] \left(\sum_{k=1}^s \frac{z^k}{k!}\right)^m.$$

The following Maple code implements a brute force calculation to verify these values and the generating functions as well.

P := proc(n, m, s) n!*coeftayl(add(z^k/k!, k=0..s)^m, z=0, n); end;
Q := proc(n, m, s) n!*coeftayl(add(z^k/k!, k=1..s)^m, z=0, n); end;

P_ex :=
proc(n, m, s)
    option remember;
    local ind, d, ms, k, res, pos;

    res := 0;

    if m = 1 then
        if n <= s then res := res + 1 fi;
        return res;
    fi;

    for ind from m^(n+1) to m^(n+1)+m^n-1 do
        d := convert(ind, base, m);
        ms := convert([seq(d[k], k=1..n)], multiset);

        for pos to nops(ms) do
            if ms[pos][2] > s then
                break;
            fi;
        od;

        if pos = nops(ms)+1 then
            res := res + 1;
        fi;
    od;

    res;
end;


Q_ex :=
proc(n, m, s)
    option remember;
    local ind, d, ms, k, res, pos;

    res := 0;

    if m = 1 then
        if n <= s then res := res + 1 fi;
        return res;
    fi;

    for ind from m^(n+1) to m^(n+1)+m^n-1 do
        d := convert(ind, base, m);
        ms := convert([seq(d[k], k=1..n)], multiset);

        for pos to nops(ms) do
            if ms[pos][2] > s then
                break;
            fi;
        od;

        if pos = nops(ms)+1 and nops(ms) = m then
            res := res + 1;
        fi;
    od;

    res;
end;

I was not able to find OEIS entries for these.

Answer 2

Include-Exclude with the sets $A_j = \{ \text{ box j has more than s balls } \}, \,\, j=1,2,\dots,m$. The number of ways is

$$ w := \left| \left(\bigcup_{j=1}^m A_j \right)^C \right| = \sum_{r=0}^m (-1)^r \binom{m}{r} |\{\text{ boxes 1,2,...,r have more than s balls }\}| $$

because for an intersection of size $r$ it doesn't matter which particular $j$'s are in it. To calculate the ways where boxes $1,2,\dots, r$ have more than $s$ balls (i.e. at least $s+1$ balls), we break it into cases of how many balls in total go to the boxes $1,2\dots, r$ (call that $u$) and use the associated Stirling numbers of the second kind ${a \brace b}_{\geq c} = $ number of ways to partition a set of size $a$ into $b$ parts where each part has at least $c$ elements. Notice that this doesn't account for the order of the parts (i.e. bins) so we must multiply by the way they can be ordered. Luckily, since the balls are distinguishable, every part is different and this is just $r!$ when there are $r$ parts.

So the answer is

$$ w = \sum_{r=0}^m (-1)^r \binom{m}{r} \sum_{u=r(s+1)}^n \binom{n}{u} r! {u \brace r}_{\geq s+1} (m-r)^{n-u} $$