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.