We will solve the problem step by step:

Problem Definitions:

1. The tree has vertices with degrees either $1$ or $5$.
2. There are $30$ vertices of degree $5$.
3. Each degree-5 vertex is either:
   • Weak: Connected to 4 degree-1 vertices and 1 degree-5 vertex, or
   • Strong: Connected to 5 degree-5 vertices.
4. We need to find:
   1. The total number of vertices.
   2. The number of strong vertices.
   3. The greatest path length between two weak vertices.

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Step 1: Total Number of Vertices

Let:

• $s$ = number of strong vertices,
• $w$ = number of weak vertices,
• $n_1$ = number of degree-1 vertices.

The number of vertices of degree $5$ is: $s + w = 30.$

Each weak vertex is connected to $4$ degree-1 vertices. Thus: $n_1 = 4w.$

Using the Handshaking Lemma ($2E = \sum \text{degree of all vertices}$), the total number of edges $E$ is: $E = \frac{n_1 + 5(s + w)}{2}.$

Since $n_1 = 4w$, we substitute: $E = \frac{4w + 5(s + w)}{2} = \frac{4w + 5s + 5w}{2} = \frac{9w + 5s}{2}.$

Now substitute $s = 30 - w$: $E = \frac{9w + 5(30 - w)}{2} = \frac{9w + 150 - 5w}{2} = \frac{4w + 150}{2} = 2w + 75.$

The total number of vertices $V$ is: $V = n_1 + s + w = 4w + s + w = 4w + 30.$

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Step 2: Number of Strong Vertices

The degree-5 vertices (30 total) are either strong or weak. From the connections:

• A strong vertex is connected to 5 degree-5 vertices, consuming 5 edges.
• A weak vertex is connected to 4 degree-1 vertices and 1 degree-5 vertex, consuming 5 edges.

The balance of edges can be used to find $s$. The equation becomes: $5s + 5w = 2E = 2(2w + 75).$

Substitute $E = 2w + 75$ into $5(s + w)$: $5s + 5w = 4w + 150.$

Rearrange to isolate $s$: $5s = 4w + 150 - 5w = -w + 150.$

Thus: $s = \frac{-w + 150}{5}.$

We also know $s + w = 30$, so substitute $s = 30 - w$ into the above: $30 - w = \frac{-w + 150}{5}.$

Multiply through by 5: $150 - 5w = -w + 150.$

Simplify: $150 - 150 = 5w - w.$

Thus: $w = 24.$

Substitute $w = 24$ into $s + w = 30$: $s = 6.$

So:

• $w = 24$,
• $s = 6$.

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Step 3: Greatest Path Length Between Two Weak Vertices

The greatest path length between two weak vertices occurs in a path with the maximum possible connections via strong vertices.

Case Analysis:

• Each strong vertex connects to at most $2$ other strong vertices.
• Strong vertices form small chains or stars, limiting the total path length.

Let the longest chain of strong vertices be $k$. The path from one weak vertex to another through $k$ strong vertices is: $\text{Path length} = 4 + k + 4 = k + 8.$

The greatest $k$ is determined by the $s = 6$ strong vertices: $k = 2 \quad (\text{since each strong vertex connects to at most 2 strong vertices}).$

Thus: $\text{Greatest path length} = 2 + 8 = 10.$

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Final Answers:

1. Total number of vertices: $4w + 30 = 4(24) + 30 = 126.$
2. Number of strong vertices: $6.$
3. Greatest path length between two weak vertices: $10.$

Let me know if you'd like further details or additional clarifications!

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Follow-Up Questions:

1. How would the results change if each strong vertex could connect to 3 other strong vertices?
2. What would the total number of edges be in the tree?
3. Can we generalize the formula for the number of vertices in such a tree?
4. How would the problem change if weak vertices could connect to 5 degree-1 vertices?
5. What is the average degree of vertices in the tree?

Tip:

Always verify your solution using the Handshaking Lemma, as it ensures consistency in edge and degree calculations.