I have the following (probably classic) combinatorics problem:
There are $n$ bins that can hold $k$ items each, and a total of $r = n\,k$ items. The items have weights $w_1 > w_2 > \dots > w_r$. The weights are i.i.d. drawn from some continuous distribution with c.d.f. F and p.d.f. f > 0
The actual problem is driven by auctions where a number of bidders are allocated to $n$ auctioneers so that $k$ exactly bidders are in the same auction and their bids are i.i.d. drawn from $F$ (the distribution should be the same in all auctioneers).
What are the probabilities that:
(i) Items $1$ and $2$ are in the same (any) bin but item $3$ is in one of the other $n-1$ bins?
(ii) items $1$,$2$, and $3$ are in the same (any) bin but item $4$ is in one of the other $n-1$ bins?
(...) Items $1$, $\dots$, $k-1$ are in the same bin but item $k$ is in one of the other $n-1$ bins?
I know that the problem is solved by considering all the possible free positions for the items $(i+1) \dots (i+k)$ when looking at the first $i$ items being in the same bin.
Given that the order matters here (they get in sorted order since for $n=2$, $k=3$, if items $1$ and $2$ are in the same bin, then the other bin can have $\{3,4,5\}$ or $\{3,4,6\}$ or $\{4,5,6\}$ i.e. $\frac{(nk-i)!}{(nk-i-k+i)!}$