Show that $N_j(n,k,r)=\binom{k}{r} \sum_{i=r}^{k} (-1)^{i-r} \binom{k-r}{i-r} \frac{n!}{(j!)^i (n-ij)!} (k-i)^{n-ij}$ The question asks
let $N_j(n,k,r)$ be the number of distributions of $n$ distinguishable balls into $k$ distinguishable urns, so that $r$ urns are occupied each by $j$ balls. Show (above expression)
$$
N_j(n,k,r)=\binom{k}{r} \sum_{i=r}^{k} (-1)^{i-r} \binom{k-r}{i-r} \frac{n!}{(j!)^i (n-ij)!} (k-i)^{n-ij}
$$
I'm assuming there is some need to use the number of divisons of a finite set with $n$ in $r$ ordered subsets, i.e.
$$C(n,k_1,...,k_{r-1})=\binom{n}{k_1,...,k_{r-1}}=\binom{n}{k_1}\binom{n-k_1}{k_2}...\binom{n-k_1-...-k_{r-2}}{k_{r-1}}=\frac{n!}{k_1!...k_{r-1}!k_r!}$$ where $k_r=n-k_1-...-k_{r-1}$.
It looks like with $k_1=k_2=...=k_{r-1}=j$ in the summation, the coefficient becomes
$$C(n,j,...,j)=\frac{n!}{(j!)^i(n-ij)!}$$
It looks like there is a relevant theorem, though not expanded. Let $A_1,...,A_n$ be exchangable sets of finite set $\Omega$, then the number $N_{n,k}$ of elements $\Omega$ contained in $k$ among $n$ subsets is given by
$$N_{n,k}=\sum_{r=k}^{n}(-1)^{r-k}\binom{r}{k}\binom{n}{r}v_r=\binom{n}{k}\sum_{r=k}^{n}(-1)^{r-k}\binom{n-k}{r-k}v_r$$
where $v_r=\sum N(A_{i_1},...,A_{i_r})$ is the number of distributions of $n$ distinguishable balls into the remaining $r$ distinguishable urns, for a selection of indices $\{ i_1,...,i_r\}$ for the sets $\{ 1,...,n\}$.
It looks like the problem can be solved from here, but i'm not entirely sure how to set up the sum. Any actual help would be appreciated.
 A: There are $n!$ permutations of the balls. For each of them we put the first $j$ balls into the first urn, the second $j$ balls into the second urn, and so on, until we’ve put $j$ balls into each of the first $i$ urns. We don’t care about the order of the balls within each of the first $i$ urns, so permutations of the balls that have the same balls in each of the first $i$ blocks of $j$ balls are equivalent. We also don’t care about the order of the remaining $n-ij$ balls, because we’re going to place each of them individually, so there are $\frac{n!}{(j!)^i(n-ij)!}$ distinguishable classes of permutations of the balls.
Now we put each of the remaining $n-ij$ balls into one of the remaining $k-i$ urns; this can be done in $(k-i)^{n-ij}$ ways. Thus,
$$\frac{n!}{(j!)^i(n-ij)!}(k-i)^{n-ij}\tag{1}$$
is the number of distributions of $n$ distinguishable balls into $k$ distinguishable urns in such a way that each of the first $i$ urns gets $j$ balls. Clearly it is also the number of distributions of $n$ distinguishable balls into $k$ distinguishable urns in such a way that each of any designated set of $i$ urns gets $j$ balls; it was just easiest to explain the first factor in terms of the first $i$ blocks of $j$ balls. Thus, $(1)$ is the $v_r$ of your final displayed expression, and you have your result.
A: This may also be done using exponential generating functions. We have
from first principles using EGFs
$$n! [z^n] {k\choose r} \left(\frac{z^j}{j!}\right)^r
\left(\exp(z) - \frac{z^j}{j!}\right)^{k-r}
\\ = n! [z^n] {k\choose r} \left(\frac{z^j}{j!}\right)^r
\sum_{q=0}^{k-r} {k-r\choose q} (-1)^q \frac{z^{qj}}{(j!)^q}
\exp((k-r-q)z)
\\ = n! [z^n] {k\choose r} \left(\frac{z^j}{j!}\right)^r
\sum_{q=r}^{k} {k-r\choose q-r} (-1)^{q-r} \frac{z^{(q-r)j}}{(j!)^{q-r}}
\exp((k-q)z)
\\ = n! [z^n] {k\choose r}
\sum_{q=r}^{k} {k-r\choose q-r} (-1)^{q-r} \frac{z^{qj}}{(j!)^q}
\exp((k-q)z)
\\ = {k\choose r}
\sum_{q=r}^{k} {k-r\choose q-r} (-1)^{q-r}  [z^{n-qj}]
\frac{n!}{(j!)^q} \exp((k-q)z)
\\ = {k\choose r}
\sum_{q=r}^{\min(k, \lfloor n/j \rfloor)}
{k-r\choose q-r} (-1)^{q-r}
\frac{n!}{(j!)^q \times (n-qj)!} (k-q)^{n-qj}.$$
This is the claim. Here we have used the combinatorial class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SEQ}_{=r}(\textsc{SET}_{=j}(\mathcal{Z}))
\textsc{SEQ}_{=k-r}(\textsc{SET}_{\ne j}(\mathcal{Z})).$$
