Given the following template of a simple cubic bipartite graph:
$\hskip1.7in$
Missing edges shall be drawn from the top nodes to the bottom nodes. No loops and multiedges allowed.
How many distinct simple cubic bipartite connected graphs can be drawn and which of them are not hamiltonian?
I tried to come up with a pure combinatorial approach, 6 people pick 2 non-identically colored balls from a set of two copies of 6 colored balls, but this gets messy pretty soon and I'm unsure about the effects of graph symmetries...
I can do the combinatorics part, but it's not encouraging. Each of the nodes we have to connect must have two more neighbor if the graph is to be cubic. Imagine that each node has two connectors, so that we have $6$ pairs of connectors on the top and $6$ on the bottom. Then we need to connect each of the bottom $12$ connectors to on of the top $12$ connectors, subject to the condition that two connectors from the same pair on the bottom cannot be connected to both connectors of one pair on the top. I will refer to this as connecting a pair.
We can do this with the principal of inclusion and exclusion. There are $12!$ ways to make the connections, and we must subtract the number of ways that a pair is connected. There are $6$ ways to choose the pair on the bottom and another $6$ ways to choose the pair on the top. There are $2$ ways to make the actual connection, since the first connector on the bottom can be connected to the first or second on the top. The remaining $10$ connectors on the bottom can be connected to the remaining $10$ on the top in $10!$ ways, giving a total of $$\binom61^210!$$ ways.
Of course, any configuration with two connected pairs has been subtracted twice, so we must add those back in, and so on. Continuing the reason above gives a total of $$\sum_{k=0}^6(-1)^k\binom6k^2k!2^k(12-2k)!=278,323,200$$ The $k!$ term comes because once we have chosen $k$ pairs on the bottom and $k$ on the top, there are $k$ ways to decide which to connect to which.
This overstates the answer by a factor of $2^{12}$ because we can interchange each pair of connectors. Dividing by $2^{12}$ gives $$67,950$$ labelled cubic bipartite graphs.
Surely, many of these are isomorphic, but I don't see a way to modify the argument above to count the isomorphism classes, let alone to produce them.
EDIT
nauty counts $1140$ isomorphism classes of connected, cubic bipartite graphs with $10$ nodes in each bipartition set. If this problem is important, one way to start would be to generate those $1140$ graphs with nauty, and test to them to see if they are in the class contemplated in this problem.
ANOTHER EDIT
nauty also has a heuristic function to test if a graph is hamiltonian. I ran it on the $1140$ graphs; $1139$ have hamilton cycles, and one timed out. It's possible that this is a false negative, but I doubt it. I pumped up the limit on the number of tries from $1$ to $100$ and got the same result. Still, it would take a custom program to actually answer your question and to determine if the possibly non-hamiltonian graph is one of the ones you're interested in.
COMPUTER RESULTS
I wrote a python script to process the $1140$ graphs. It computed the second neighborhood of each vertex $v$, that is, the set of neighbors of $v$'s neighbors, excluding $v$ itself. If a graph has two vertices with second neighborhoods of cardinality $6$, and those neighborhoods are disjoint, we can take one of the vertices as the top vertex in the drawing, and one as the bottom vertex. $963$ of the $1140$ graphs met this criterion, and $962$ of them have hamiltion cycles.
The one where a hamilton cycle was not found was the simplest one. In the diagram join the two leftmost free nodes on the bottom to the two leftmost free nodes on the top, and similarly for the pairs in the middle and the pairs on the the right.
ADDENDUM
I don't know why I can't let this go, but it turns out to be easy to prove that the anomalous graph doesn't have a hamiltonian cycle. If it has a hamiltonian cycle, we can color the edges of the cycle alternately red and green, since the graph has an even number of vertices. Then we can color all the remaining edges blue, so there is a $3$-coloring of the edges with a two-colored hamiltonian cycle.
Here is a picture of the graph:
Consider a $3$-coloring of the edges. Without loss of generality, we may assume that AH is colored red, AB and HI are colored green, and AE and HJ are colored blue, as shown. One of JK and JL must be colored red, and one of IK and IL must be colored red. Therefore, one of KI and KJ is red, and KQ is not red. Similarly, LQ is not red, and therefore QT is colored red. Similarly RT must be green and ST must be blue.
Now, no two-color chain can be a hamiltonian cycle. For example, a red-green chain starting at A will not include AE or ST, so it will never get to the right-hand portion of the diagram.