Number of cycles in complete graph How many number of cycles are there in a complete graph?
Is there any relation to Symmetric group?
 A: We can use some group theory to count the number of cycles of the graph $K_k$ with $n$ vertices.  First note that the symmetric group $S_k$ acts on the complete graph by permuting its vertices.  It's clear that you can send any $n$-cycle to any other $n$-cycle via this action, so we say that $S_k$ acts transitively on the $n$-cycles.  The orbit-stabilizer theorem states that the order of a group acting transitively on a set, is the product of the size of the set and the size of the subgroup stabilizing an element of the set.  In this case, we can stabilize an $n$-cycle by permuting the $k-n$ vertices not involved in the cycle, and then permuting the $n$ vertices in the cycle in a way that preserves the cycle.  This gives us that the cycle stabilizer has size $(k-n)!\cdot 2n.$  Now we have $$|S_k| = (\text{number of n-cycles})((k-n)!\cdot 2n). $$ hence the number of $n$-cycles is $\frac{k!}{(k-n)!\cdot 2n}$.  The total number of cycles can be computed as a sum: $$\sum_{i=3}^k \frac{k!}{(k-i)!\cdot 2i}.$$ I'm not sure whether this sum simplifies.
Here the group theory doesn't add much to the counting, over the usual overcounting-and-dividing solution to this type of problem.  However it demonstrates that this technique is a special case of a more general result, and gives a concrete example with which to understand it. 
A: From $K_n$ lets pick a number $p$, $p \le n$ where $C_p$ is a cycle of length $p$.
Let $2\to 3\to 4\to 5\to 1\to\dots\to (p-1)\to 2$ be any arbitrary $C_p$. (note that the first and last vertex are same).
Now we fix the first vertex(which is also the last vertex since it is a cycle) like:
$$
2\to [3\to 4\to 5\to 1\to\dots\to (p-1)]\to 2
$$
We have $(p-1)!/2$ different cycles if we permute the section within the square brackets '$[]$' (circular permutation).
Now, from $K_n$ we can choose $p$ vertex in $\dbinom np$ ways and each of the $p$ vertex has $(p-1)!/2$ cycles.
Therefore the total number of cycles in $K_n$ is 
$$
\sum_{p=3}^{n} \binom np \frac{(p-1)!}{2}.
$$
A: In a complete graph, every choice of n vertices is a cycle, so if the graph has k vertices, then there is $\sum_{n=3}^{k} {k \choose n}$, which is equal to $  \dfrac{-k^2}{2}-\dfrac{k}{2}+2^k-1$. As for the symmetric group, I'm pretty sure that it is the automorphism group for the complete graph of the same size.
