Suppose:
- $i$ boxes contain red rocks.
- $j$ boxes contain black rocks.
- $s$ red rocks are in boxes.
- $t$ black rocks are in boxes.
(No, I'm not trying to write a Dr Seuss book.)
With these restrictions, there are ${26 \choose {i,j,26-i-j}}$ ways to choose the boxes for the rocks (note: this is the multinomial coefficient). Once we've decided on which boxes red rocks belong to, we need to partition them into boxes. Let $N_{n,k}$ be the number of ordered partitions of $n$ into $k$ non-zero parts. Then the number of ways of putting the red rocks into their chosen boxes is $N_{s,i}$. Similarly, the number of ways of putting the black rocks into their chosen boxes is $N_{t,j}$.
Hence, if we sum this over all valid $i,j,s,t$ we get the result \[\sum_{i=0}^{15} \sum_{j=0}^{15} \sum_{s=0}^{15} \sum_{t=0}^{15} {26 \choose {i,j,26-i-j}} N_{s,i} N_{t,j}.\]
(Technically, we need the upper bound on $j$ to be $\mathrm{min}(15,26-j)$, but if we assume that all multinomial coefficients with a negative parameter are $0$, then it's fine.)
This can be computed in GAP by:
Multinomial:=function(n,L) local x,i; if(ForAny(L,i->i<0)) then return 0; fi; if(Sum(L)<>n) then return 0; fi; if(n=0) then return 1; fi; x:=Factorial(n); for i in L do x:=x/Factorial(i); od; return x; end;; num:=0; for i in [0..15] do for j in [0..15] do for s in [0..15] do for t in [0..15] do num:=num+Multinomial(26,[i,j,26-i-j])*NrOrderedPartitions(s,i)*NrOrderedPartitions(t,j); od; od; od; od; Print(num,"\n");
which gave the result 38996875553354247211
.
As a quick check of this number, we can act on the set of arrangements (by permuting box indices) with the group $\langle (1,2,\ldots,p)\rangle$ for prime $p$ in the range $15. Orbits have size $p$ unless the first $p$ boxes are empty (since $p>15$).
The number of arrangements with the first $p$ boxes empty is listed below (computed using an edited version of the above code), and the congruences match up.
p nr arrangements with boxes 1..p empty 17 25995200573833414041 19 3229548304818580291 23 1017952468781
As another check, we can act on the set of arrangements with the group $\langle (1,2,\ldots,13)(14,15,\ldots,26)\rangle$. Non-identity permutations in this group have no fixed points and have order $13$. Thus, orbits have size $13$, unless they are one of the ones listed below:
............. ............. ............. rrrrrrrrrrrrr ............. bbbbbbbbbbbbb rrrrrrrrrrrrr ............. rrrrrrrrrrrrr bbbbbbbbbbbbb bbbbbbbbbbbbb ............. bbbbbbbbbbbbb rrrrrrrrrrrrr
(note: we must have only one copy of a symbol in each box, since $2 \times 13>15$). This is consistent with $38996875553354247211 \equiv 7 \pmod {13}$.