Total node = (Degree of Node) * (No of Node of that degree) + 1
Total Node = Leaf Node + Internal Node

Let number of nodes of degree 2 is n and number of nodes of degree 3 is m,

Total node = 2n + 3m + 1
Total node = 9 + (m + n)

Both are equal so equate them,
2n + 3m + 1 = 9 + (m + n)
n + 2m = 8

Solution 1 : n = 2, m = 3
Solution 2 : n = 4, m = 2
Solution 3 : n = 6, m = 1
Solution 4 : n = 8, m = 0

Solution 1,2,3 are feasible but out of these three we have to select solution with max number of nodes i.e. solution 3.
Internal nodes are 7 leaf nodes are 9, Total 16 nodes.

Why not solution 4 ??
in question there is one more line "all paths from root to the leaves have the same length." With solution 4 this property violates. In complete binary tree with d depth max no of nodes will be 2^d-1. By putting d = 4 we wont get 16. i.e. max 15 possible. For 16th we have to go to next level and then "all paths from root to the leaves have the same length." violated.

In trees, #edges = #nodes – 1 In trees, degree of a node = #children of that node By handshaking lemma, #edges = For node i = 1 to i = n Σ( #children of a node i)

#edges = degree of a node * #nodes with that degree