The Lowest Level PL

Summary

Treat the two \(R\)-cubes as a single block \(RR\). The remaining multiset is \(\{RR, G,G, B,B, E,E,E\}\) of size 8 with multiplicities \(G:2, B:2, E:3\). Thus \[ N(A_R)=\frac{8!}{2!\,2!\,3!}. \] Compute: \begin{align*} 8!&=40\,320,\\ 2!\,2!\,3!&=2\cdot2\cdot6=24,\\ N(A_R)&=\frac{40\,320}{24}=1\,680. \end{align*} By symmetry the same value holds for \(N(A_G)\) and \(N(A_B)\). Hence \[ \sum_X N(A_X)=3\cdot1\,680=5\,040. \]