Given Bipartite Graph $K(a,b)$ how many cycles in size 4 can be?
As we know $K(a,b)$ is a full Bipartite Graph meaning that for every $vi\in a$ there is an edge to every $v\in b$.
To get a cycle at size 4 I will need to pick one $vi\in a$ that will be connected via edges to $vi,vj\in b$.
Then to both those $vi,vj\in b$, I will pick another, different from the one above, $vj\in a$ that they will connect via edges.
In order to calculate it, I've tried to use combinatorics:
To choose the first $vi\in a$ I will have $a\choose1$ options
To choose the $vi,vj\in b$ I will have $b\choose2$
And to choose the $vj\in a$ I will only have $(a-1)\choose1$.
So the total possibilities that I have will be $a\choose1$ $\cdot$ $b\choose2$ $\cdot$ $(a-1)\choose1$
Is it correct to say that?
Let's first choose the vertices, a distinct pair from the part of size $a$ and a distinct pair from the part of size $b$. (We consider the two parts distinguishable even if they have the same size.)
Those four vertices can be connected in a cycle in only one way, although the cycle could be "oriented" or directed in two ways. Therefore the number of cycles is:
$$ \binom{a}{2} \binom{b}{2} $$
This differs from the count made by the OP. A small example illustrates that the method used there double counts because the same choice of a pair of vertices in the part of size $a$ occurs twice (in either order). E.g. take the complete bipartite graph $K_{2,2}$, which has only one $4$-cycle.