They have played with words quite well **up to** should be in bold

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64 votes

@rupesh17 Your Generalized solution is not correct. For 4 unlabeled vertices we have 11 simple graphs (not 7). See below references.

https://mathworld.wolfram.com/SimpleGraph.html

https://garsia.math.yorku.ca/~zabrocki/math3260w03/nall.html

As per my current understanding no general solution for unlabeled vertices simple graph.If anyone knows please comment.

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110 votes

Also from Wikipedia:

"Some authors exclude K0 from consideration as a graph (either by definition, or more simply as a matter of convenience). Whether including K0 as a valid graph is useful depends on context."

Same thing I told in previous message.

Half-reading is misleading & dangerous.

IIT, IISc, NPTEL professors consider graphs to have at least one vertex. You can follow that.

1

This is a screenshot from the very famous graph theory book by Reinhard Diestel, often called as "Diestel's graph theory book" which is considered as a standard textbook and followed by many people all over the world.

Now, we can see vertex set can be empty and according to this 8 should be the correct answer provided graph is unlabeled. So,”theoretically” both 7 and 8 are the "possible" answers for an unlabeled graph. It should not be like that: 7 is correct and 8 is not correct though it is another case that $8$ is not given as the option.

To avoid ambiguity in the questions, it would be very helpful if more details would be added in the questions. It might create long questions but chances of ambiguity will be less.

Graph Theory is considered as a young subject and so, terms should be properly defined in the question. For example, a complete graph is defined as a graph whose vertices are pairwise adjacent, while a clique is a set of pairwise adjacent vertices in a graph. Many authors use the terms interchangeably. And, there are other examples too. So, to avoid ambiguity, it would be much helpful if things are defined properly and as much as details should be added in the question.

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edited
Aug 7, 2022
by Deepak Poonia

That’s well written.

It depends on the author whether they allow empty vertex set or not in the graph definition. Particularly, in IISc CSA Graph Theory course also, Diestel is followed, But Prof doesn’t allow empty vertex set in the definition of Graph.

Conclusion: 7, 8 both are correct answers, But since 8 is Not given in the options, so, 7 is unambiguously correct answer.

3

50 votes

Answer: C

The number of max edges a simple graph can have is $\frac{n×(n−1)}{2}$.

So, for a graph with $3$ nodes the max number of edges is $3.$

Now there can be $0$ edges, $1$ edge, $2$ edges or $3$ edges in a $3$ node simple graph.

So the total number of unlabeled simple graphs on $3$ nodes will be $4.$

Similarly for two node graph we have option of $0$ or $1$ edge and for one node graph we have option of $0$ edge.

So the total number of simple graphs upto three nodes $=4+2+1=7.$

The number of max edges a simple graph can have is $\frac{n×(n−1)}{2}$.

So, for a graph with $3$ nodes the max number of edges is $3.$

Now there can be $0$ edges, $1$ edge, $2$ edges or $3$ edges in a $3$ node simple graph.

So the total number of unlabeled simple graphs on $3$ nodes will be $4.$

Similarly for two node graph we have option of $0$ or $1$ edge and for one node graph we have option of $0$ edge.

So the total number of simple graphs upto three nodes $=4+2+1=7.$

35 votes

With n number of nodes, max edges are possible $\frac{n(n-1)}{2}$ in a simple graph. each edge has two choices, it'll be taken in a graph or it won't be taken. so total no of graphs are possible

$2^{\frac{n(n-1)}{2}}$.

in the question, it is asked upto three nodes:

with 1 node total graph possible = 1

with two nodes graph possible = 2

with three nodes, possible graphs= $2^{\frac{3(3-1)}{2}} = 2^3 = 8$ but there are 3 non distinct graphs which are isomorphic to each other, so with n=3 nodes total graphs possible = 8 but total unique graph possible = 4

upto 3 nodes $1 + 2 +4 = 7 $hence answer is (C)

$2^{\frac{n(n-1)}{2}}$.

in the question, it is asked upto three nodes:

with 1 node total graph possible = 1

with two nodes graph possible = 2

with three nodes, possible graphs= $2^{\frac{3(3-1)}{2}} = 2^3 = 8$ but there are 3 non distinct graphs which are isomorphic to each other, so with n=3 nodes total graphs possible = 8 but total unique graph possible = 4

upto 3 nodes $1 + 2 +4 = 7 $hence answer is (C)

0

14 votes

A graph is an ordered pair (V,E). So, given a set V, graph will be different if E is different. Lets see how many different E we can get when |V| = 3.

For |V| = 3, we can have |E| = 0, 1, 2 or 3. So, 4 possible graphs.

Now, the question asks for |V| upto 3. So, we have to consider |V| = 2 and |V| = 1 also. When |V| = 2, we can have |E| = 1 or 0, so 2 possibilities. For |V| = 1, |E| can be only 0 and hence only one possibility. So, total number of possibilities is $$4 + 2 + 1 = 7$$.

For |V| = 3, we can have |E| = 0, 1, 2 or 3. So, 4 possible graphs.

Now, the question asks for |V| upto 3. So, we have to consider |V| = 2 and |V| = 1 also. When |V| = 2, we can have |E| = 1 or 0, so 2 possibilities. For |V| = 1, |E| can be only 0 and hence only one possibility. So, total number of possibilities is $$4 + 2 + 1 = 7$$.