How many graphs on n labeled vertices exist which have at least (n2 – 3n)/2 edges ?
Let there be total number of edges of a graph to be formed be e and total number of vertices be n.
So the total number of ways a graph can be formed having exactly e edges and v vertices is given by the total number of ways we can select e edges among v edges
C (v, e)
Now, here e = (n2 – 3n)/2 and v = n(n-1)/2 (maximum number of edges in a simple graph).
So to select a edge we can do that in C (v, (n2 – 3n)/2) ways.
Since minimum no. of edges to be selected for a graph are (n2 – 3n)/2.
So total number of graphs possible will be :
C(v, e) + C(v, e+1) + C(v, e+2) +……………+ C(v, v).
C(v, v-e) + C(v, v-(e+1)) + C(v, v-(e+2)) +……………+ C(v, v-v).
Since v – e = n(n-1)/2 – (n2 – 3n)/2 = n
C(v, n) + C(v, n-1) + C(v, n-2) +……………+ C(v, 0).
Solving this we will get
This solution is contributed by Namita Singh.