Let $a, b \geq 3$ and let $G$ consist of two complete subgraphs $G_a$ and $G_b$ that share exactly two vertices. How many Hamiltonian cycles are in $G$?
I was close! As answered below, $(K_a - 2)! * (K_b - 2)!$
The $-2$ is because the $2$ connecting points are somewhat fixed to bridge the two
Let $u$ and $v$ be the two shared vertices. Any Hamilton cycle (viewed as a subgraph and not as a walk) would be the union of a $(u,v)$-Hamilton path in $K_a$ and a $(u,v)$-Hamilton path in $K_b$. Using the Multiplaction Rule, the number of such cycles is $(a-2)!\cdot (b-2)!$