Using the Pr¨ufer correspondence, for n ≥ 10, count the number of trees with vertex set [n] that have maximum degree 3 and exactly six leaves. (Hint: Start from finding out how many vertices of degree 3 such a tree has.)

The Prüfer correspondence is a method used in combinatorics to establish a one-to-one correspondence between the set of labeled trees and the set of sequences, both of size n-2.

To solve this problem, we first need to determine how many vertices of degree 3 such a tree has.

1. Since the tree has exactly six leaves, and leaves are vertices of degree 1, we know that there are n-6 vertices that are not leaves.

2. A tree with n vertices has n-1 edges. Each edge contributes to the degree of two vertices. Therefore, the sum of the degrees of all vertices is 2(n-1).

3. The sum of the degrees of the leaves is 6, since each leaf has degree 1. Therefore, the sum of the degrees of the non-leaf vertices is 2(n-1) - 6.

4. Each non-leaf vertex has degree 3. Therefore, there are (2(n-1) - 6) / 3 non-leaf vertices.

5. Since the number of non-leaf vertices must be an integer, n must be of the form 9k+4 for some integer k.

Now, we need to count the number of such trees.

1. The Prüfer sequence of a tree with n vertices is a sequence of n-2 numbers, each of which is a label of a vertex.

2. In our case, the labels of the non-leaf vertices can appear in the Prüfer sequence. Each non-leaf vertex can appear up to 2 times, because its degree is 3 and one of the edges is not counted in the Prüfer sequence.

3. Therefore, the number of such trees is the number of sequences of length n-2, where each number is a label of a non-leaf vertex and each non-leaf vertex appears at most 2 times.

4. This is equivalent to the number of ways to distribute n-2 balls into (2(n-1) - 6) / 3 boxes, where each box can contain at most 2 balls.

5. This can be calculated using the formula for distributing balls into boxes with restrictions, which is a topic in combinatorics.