Determine the number of Hamiltonian cycles in K2,3 and K4,4 and the existence of Euler trails

1. Determine whether there exist Euler trails in the following graphs
2. Determine the number of Hamiltonian cycles in K2,3 and K4,4

My approach:

A1. After observing graph 1, 8 vertices (boundary) have odd degrees.

It is contradictory to the definition (exactly 2 vertices must have odd degree).

In graph 2, there exists euler trails because exactly 2 vertices (top left- outer region and top right- outer region) have odd degrees.

A2. Hamiltonian cycle: contains every vertex one and only one time or proving by Dirac's theorem.

Following the Dirac's theorem:

For K2,3, number of vertices, n= 5, n/2= 2.5

For 2 vertices, deg (v)= 3; for the other 3 vertices, deg (v) = 2 (which is less than 2.5)

To satisfy Dirac's condition, for every vertex, v, deg (v)>=n/2. Hence, no hamiltonian cycle.

For K4,4, number of vertices, n=8, n/2 =4 For all the veritces, deg (v)= 4;

Hence, hamiltonian cycle exists.

But I am not sure how to find the total number of hamiltonian cycles.

For $K_{4,4}$ you can refer to this question.

For $K_{2,3}$, Suppose that you have an Hamiltonian cycle $(v_1v_2v_3v_4v_5)$.

Your graph is composed of two independent set of vertices $A$ and $B$, with $\vert A\vert=3$ and $\vert B\vert = 2$. Without loss of generality, we can suppose that $v_1\in A$ (otherwise look at the the cycle $(v_2v_3v_4v_5v_1)$).

Then $$v_1\in A \Rightarrow v_2\in B \Rightarrow v_3\in A \Rightarrow v_4\in B \Rightarrow v_5\in A \Rightarrow v_1\in B $$ Hence a contradiction. In general, for any $n\neq m$, $K_{n,m}$ is not Hamiltonian (but Dirac's theorem is not a proof)