How many possible cycles are there in a clique of four vertices? Here is the graph:

The answer should be 48 cycles. I know that you can have a circuit of 3 or 4 vertices. I don't know how to show my steps from there.
 A: I wouldn't say that this is counting cycles, exactly. 
We can get $48$ if we count sequences of $3$ or $4$ distinct vertices $(v_1, v_2, \dots, v_n)$, $n \in \{3,4\}$; presumably, in general, there's the requirement that $v_i v_{i+1}$ should be an edge for all $i$, as should $v_n v_1$, but in this particular graph that's always true.
If this is what we're counting, then:


*

*For sequences of length $4$, there are $4!$ ways to choose $(v_1, v_2, v_3, v_4)$, which is a permutation of $(1,2,3,4)$.

*For sequences of length $3$, there are $4$ ways to choose which vertex is left out, and then $3!$ ways to permute the remaining vertices.


So the answer we get is $4! + 4\cdot 3! = 24 + 24 = 48$.
I hesitate to call these cycles because, even though each such sequence $(v_1, v_2, \dots, v_n)$ describes a cycle in this graph, any cycle has multiple descriptions along these lines. For example, I would consider $(1,2,3,4)$ to describe the same cycle as $(3,4,1,2)$ or $(4,3,2,1)$, because the edges used are all the same. So by my definition, there would only be $7$ cycles: $3$ cycles of length $4$ (choose which two of $\{1,2,3\}$ are adjacent to vertex $4$, which uniquely determines the rest of the cycle) and $4$ cycles of length $3$ (choose which vertex is left out).
