On $n$ nodes, we have $2^{n(n-1)/2}$ graphs. Asymmetric graph is a graph that has only trivial automorphism. We known that asymptotically almost all finite graphs are asymmetric. Therefore, in the limit, the ratio of asymmetric graphs approaches 1. 


However, I did not find any reference that provides lower bound on the number of asymmetric graphs on $n$ nodes. What is known about the density of asymmetric graphs as a function of the number of nodes $n$?


**EDIT 1-3-2014** Thanks to both answers. I received two estimates of the number of symmetric graphs. I'm still hoping for a better tight asymptotic upper-bound such as $O(f(n))$ where $n= |V(G)|$ or even better a tight asymptotic lower-bound on the number of asymmetric graphs $\omega(g(n))$.



How dense is the set of asymmetric graphs?
Almost all non-asymmetric graphs have exactly one non-trivial automorphism, namely a transposition swapping two vertices. So, an accurate estimate of their number is
obtained by taking an arbitrary graph with one vertex less, choosing a vertex $v$, adding a new vertex $w$ with the same neighbours as $v$, then either joining or not joining $v$ to $w$. For labelled graphs, if $g\_n=2^{\binom n2}$ is the number of them we have asymptotically $2ng\_{n-1}$ non-asymmetric graphs. This is a small fraction. For unlabelled graphs, divide the total by $n!$ and the non-asymmetric ones by $n!/2$.


