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How many undirected graphs (not necessarily connected) can be constructed out of a given set $V=\{v_1, v_2, \dots v_n\}$ of $n$ vertices?

  1. $\frac{n(n-1)} {2}$
  2. $2^n$
  3. $n!$
  4. $2^\frac{n(n-1)} {2} $
asked in Graph Theory by Veteran (68.9k points)
edited by | 1.8k views
The answer corresponds to graph without multiple edges and self loops i.e. simple graphs.

@Bikram Am I correct.

Shivam Chauhan

yes, generally we consider simple graphs, if nothing is mention .

@Bikram Sir
I understood the solution. But, why is

nC0 + nC1 + nC2   . . . . nCn = 2n a wrong answer ?

Since, single vertex graph is also a connected graph and there will be n such graph. Now if we take 2 vertex out of n then we form nC2 graphs and so on . .

What mistake am I committing here ?

Every selection of vertex will also include multiple graphs.

Ex you have chosen 3 nodes in nC3 ways, now u can again form many graphs from it. That's why this answer is wrong
^^

Okay, I got it. Thank you.

1 Answer

+31 votes
Best answer
With n vertices we have max possible $^{n}C_{2}$ edges in a simple graph. and each subset of these edges will form a graph, so total number of undirected  graph possible = $2^{\frac{n(n-1)}{2}}$
answered by Veteran (13.8k points)
edited by
Another way:

In a complete graph there are [n*(n-1)]/2 edges.

Any of the edge can be selected/neglected, so  2^([n*(n-1)]/2)
Can you please tell the case when I have only few vertices like only v1 ,v2,v3 ,Now this is not counted in these cases as we are just counting the total no of subsets of edges , and the case where we do not chose any edge there we shall have all the vertices isolated from each other , but how to  consider the case where we may have fewer vertices and they still remain isolated .
Sorry did nt get ur question ?? can u elaborate it ??


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