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The question provided is asking whether a full binary tree with seven vertices can exist and which explanation is correct. Let's break it down:

A **full binary tree** is a type of binary tree where every node has either 0 or 2 children (i.e., no nodes have exactly one child).

### Properties of a full binary tree:
1. **Number of vertices (nodes):**
   - A full binary tree with \( k \) internal (non-leaf) nodes has \( 2k + 1 \) total nodes (this includes both leaf and non-leaf nodes).
   - This means the total number of vertices \( V \) is always odd. In other words, if \( V \) is the total number of vertices, \( V = 2k + 1 \).

2. **Given that the question asks about seven vertices:**
   - Seven is an odd number, and the formula \( 2k + 1 \) is satisfied by \( k = 3 \), because \( 2(3) + 1 = 7 \).

This implies that a **full binary tree with seven vertices is possible**, so the correct answer is:

- **Yes. A graph with these properties can exist.**

Would you like a deeper explanation or more examples of full binary trees? Here are some related questions:

1. What is the difference between a full binary tree and a complete binary tree?
2. How many leaf nodes does a full binary tree with \( k \) internal nodes have?
3. Can a full binary tree have an even number of vertices?
4. How do you calculate the height of a full binary tree with \( n \) vertices?
5. What is the relation between the number of internal and external nodes in a full binary tree?

**Tip:** A full binary tree always has an odd number of vertices due to the \( 2k + 1 \) formula.

Can a full binary tree with seven vertices exist? If yes, provide the answer; if not, explain why.

Explore whether a full binary tree with seven vertices is possible based on graph theory principles. This problem uses the formula V = 2k + 1, explaining how seven vertices fit the properties of a full binary tree. Suitable for students in Grades 9-12.

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