Question 1176884
Let's break down this problem:

**Key Properties of Binary Trees**

* **Connected:** All vertices are reachable from any other vertex.
* **No Cycles:** There are no closed loops.
* **Degree ≤ 3:** Each vertex has at most three edges connected to it.
* **Vertices of Degree 1:** These are called leaves.

**Understanding the Relationship**

* In any tree, the sum of the degrees of all vertices is equal to twice the number of edges.
* In a tree with 'n' vertices, there are 'n-1' edges.

**Applying the Concepts**

1.  **Total Degree:**
    * With 15 vertices, there are 14 edges.
    * The sum of the degrees of all vertices is 2 * 14 = 28.

2.  **Maximizing Degree 1 Vertices (Leaves):**
    * To maximize the number of degree 1 vertices, we want to minimize the degrees of the other vertices.
    * The remaining vertices should ideally be of degree 3.

3.  **Let's use variables:**
    * Let 'x' be the number of vertices with degree 1.
    * Let 'y' be the number of vertices with degree 3.
    * We know x + y <= 15.
    * Also we know that the sum of the degrees must be 28.
    * Therefore 1 * x + 3 * y = 28.

4.  **Solving for x and y:**
    * From x + y <= 15, we get y <= 15 - x.
    * From x + 3y = 28, we get y = (28 - x) / 3.
    * We can say (28 - x) / 3 <= 15 - x
    * 28 - x <= 45 - 3x
    * 2x <= 17
    * x <= 8.5
    * Since x must be an integer, x <= 8.
    * If x = 10, then y = 6, and 10 + 6 = 16. That would not be possible, as we only have 15 vertices.
    * If x = 11, then 11 + 3y = 28. 3y = 17, and y = 17/3. This is not a whole number, so that is not possible.
    * If x = 10, then 10 + 3y = 28. 3y = 18. y = 6. 10 + 6 = 16. This is not possible.
    * If x = 8, then 8 + 3y = 28. 3y = 20, y = 20/3. This does not work.
    * If we try x = 10, then we have 10 vertices of degree 1. 5 vertices of degree 2.
    * 10 * 1 + 5 * 2 = 20, which is less than 28. This is not possible.
    * If we have 10 vertices of degree 1, we must have 5 remaining vertices. 10 + 5 = 15. The sum of the degrees is 10 + 3y.
    * We need the sum of the degrees to be 28.
    * 10 + 3y = 28. 3y = 18. y = 6.
    * 10 + 6 = 16. This is not possible.
    * Let's check 10 vertices of degree 1.
    * Leaves = 10.
    * Remaining vertices = 5.
    * 10(1) + 5(3) = 25. This is less than 28.
    * 10 vertices of degree one, and 5 vertices of degree 3.
    * If we had 10 vertices of degree 1, we would need 6 vertices of degree 3. 10+6 = 16, which is too many.
    * If we had 10 degree 1 vertices, and 5 degree 2 vertices, then we would have 20 degree total.
    * We can have 10 vertices of degree 1.

**Answer:**

The maximum number of vertices of degree one that a binary tree with 15 vertices can have is 10.