Let's solve this step-by-step.

Problem Breakdown:

We are tasked with finding the number of edges in a graph $G$ with:

1. 6 vertices.
2. 1 isolated vertex: This has degree $0$.
3. 1 pendent vertex: This has degree $1$, connected to exactly one other vertex.
4. 1 vertex of degree 2.
5. 3 vertices of degree 3 each.

Key Concept:

The sum of the degrees of all vertices in a graph equals twice the number of edges ($2|E|$), where $|E|$ is the number of edges. This is known as the handshaking lemma.

Step 1: Calculate the total degree

• The degrees of all vertices add up as follows: $\text{Total degree} = 0 (\text{isolated vertex}) + 1 (\text{pendent vertex}) + 2 (\text{vertex of degree 2}) + 3 \times 3 (\text{3 vertices of degree 3}).$ Simplifying: $\text{Total degree} = 0 + 1 + 2 + 9 = 12.$

Step 2: Use the handshaking lemma

By the lemma: $\text{Total degree} = 2|E|.$ Substitute $12$ for the total degree: $12 = 2|E| \implies |E| = \frac{12}{2} = 6.$

Final Answer:

The number of edges is 6. Hence, the correct option is: $\boxed{b) 6}.$

Would you like further clarification on any step or related graph theory concepts?

Related Questions:

1. What is the significance of the handshaking lemma in graph theory?
2. How do isolated vertices impact the total degree and edge count in a graph?
3. Can a graph with all vertices having degree 3 be planar? Why or why not?
4. What is the relationship between the sum of vertex degrees and graph density?
5. How does the presence of a pendent vertex affect the structure of the graph?

Tip:

In graph problems, always verify the total degree and confirm it matches twice the edge count for consistency!