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How many undirected graphs are possible with n vertices

  1. if graphs are not necessarily connected
  2. if they are necessarily connected


asked ago in Graph Theory by Active (1.1k points) 1 10
edited ago by | 42 views

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In the question that came in GATE 1994 , it is just asking number of simple graphs having upto 3 vertices and hence we should not bother about connectivity..

And in the question which came in GATE 2001 , it asked about the number of simple graphs of n vertices which is not necessarily connected which means it may be connected or may not be.So we have to include both types of graphs and hence connectivity is not a concern here as well..

So for both of them , the answer would be :  2(n(n-1)/2) as we can have maximum of n(n-1) / 2 edges and each edge can be selected or rejected for formation of the graph..

However if we are asked : Number of simple undirected graphs which are necessarily connected and has 'n' vertices , then this problem's complexity becomes higher as we need to take connectivity for each graph into consideration as well..

For labelled graphs of n vertices , the following recurrence is used :


Σ nCk  * k * d* 2(n-k)(n-k-1)/2    =     n * 2n(n-1)/2    where dn is the number of labelled, connected, simple graphs on n vertices.

For more ref , u may check the link :

answered ago by Veteran (88.5k points) 15 58 294
selected ago by

So for both of them , the answer would be :  2(n(n-1)/2)

Then for the GATE 1994 question, with 3 vertices, number of graph possible should be 2(3(3-1)/2) = 8, but on one of the answer it is written that

total number of unlabled simple graphs on 3 nodes will be  4.

I am sorry, but I am missing something here, I am not getting it.

Since it is unlabelled simple graphs , hence we would get same unlabelled graph for a few labelled graph and hence in this case we have to manually check which is not difficult to do for n = 3.

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