Partition n items into groups such that there is exactly one group of size k

Question

Question: Assume we have $n$ items that we seek to partition into $m$ groups. What's the probability that exactly one of the partitions has the size of $k$ (others can have less or more than $k$ items)?
More general question: Assume we have $n$ items that we seek to partition them into $m$ groups. What's the probability that exactly $t$ of the partitions have the size of $k$ (others can have less or more than $k$ items)? Is there any way to avoid complicated formulas driven by inclusion-exclusion?

Thanks

Answer 1

Starting with the combinatorial class of sets with size $k$ marked we find

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}_{\lt k}(\mathcal{Z}) +\mathcal{U} \textsc{SET}_{= k}(\mathcal{Z}) +\textsc{SET}_{\gt k}(\mathcal{Z})$$

and build the generating function

$$\exp(z) + (u-1) \frac{z^k}{k!}.$$

The probability is then given by

$$\frac{1}{A^n} n! [z^n] [u^1] \left(\exp(z) + (u-1) \frac{z^k}{k!}\right)^A.$$

To get the coefficient on $[u^1]$ we differentiate and set $u=0$, obtaining

$$\left.\frac{1}{A^{n-1}} n! [z^n] \left(\exp(z) + (u-1) \frac{z^k}{k!}\right)^{A-1} \frac{z^k}{k!}\right|_{u=0} \\ = \frac{1}{A^{n-1}} n! [z^n] \left(\exp(z) - \frac{z^k}{k!}\right)^{A-1} \frac{z^k}{k!}.$$

Extracting coefficients from this we find

$$\frac{1}{A^{n-1}} \frac{n!}{k!} [z^{n-k}] \left(\exp(z) - \frac{z^k}{k!}\right)^{A-1} \\ = \frac{1}{A^{n-1}} \frac{n!}{k!} [z^{n-k}] \sum_{q=0}^{A-1} {A-1\choose q} (-1)^q \frac{z^{qk}}{k!^q} \exp((A-1-q)z) \\ = \frac{1}{A^{n-1}} \frac{n!}{k!} [z^{n-k}] \sum_{q=0}^{\lfloor n/k\rfloor - 1} {A-1\choose q} (-1)^q \frac{z^{qk}}{k!^q} \exp((A-1-q)z) \\ = \frac{1}{A^{n-1}} \frac{n!}{k!} \sum_{q=0}^{\lfloor n/k\rfloor - 1} {A-1\choose q} (-1)^q \frac{1}{k!^q} [z^{n-(q+1)k}] \exp((A-1-q)z).$$

We thus get for our probability

$$\bbox[5px,border:2px solid #00A000]{ \frac{n!}{A^{n-1}} \sum_{q=0}^{\lfloor n/k\rfloor - 1} {A-1\choose q} (-1)^q \frac{1}{k!^{q+1}} \frac{(A-1-q)^{n-(q+1)k}}{(n-(q+1)k)!}.}$$

We can verify this formula by enumeration, which is shown below.

with(combinat);

ENUMX :=
proc(A, n, k)
    option remember;
    local res, part, psize, mset;

    res := 0;

    part := firstpart(n);

    while type(part, `list`) do
        psize := nops(part);
        mset := convert(part, `multiset`);

        if [k, 1] in mset then
            res := res + binomial(A, psize) * 
            n!/mul(p!, p in part) *
            psize!/mul(p[2]!, p in mset);
        fi;

        part := nextpart(part);
    od;

    res/A^n;
end;


X := (A, n, k) -> 
n!/A^(n-1) *
add(binomial(A-1,q)*(-1)^q/k!^(q+1) *
    (A-1-q)^(n-(q+1)*k)/(n-(q+1)*k)!,
    q=0..floor(n/k)-1);