I am looking for a formula to express the average number of full nodes (i.e. nodes having exactly $m$ children) in a $m$-ary tree having $n$ nodes, i.e., $$ \mu_{n}^{(m)} = \frac{\# \text{full nodes in all $m$-ary trees having $n$ nodes} }{\# \text{nodes in all $m$-ary trees having $n$ nodes}}$$.
As to the denominator, according to this answer, it sholud be given by $$n C_m^n = \binom{mn}{n} \frac{n}{(m-1)n + 1}$$ As to numerator, at the first glance, I tought that this answer would have provided a hint; but then I have realized that it is not the case because it gives how many trees have $n$ full nodes, but I need to know how many full nodes are in the set of all the $m$-ary trees having $n$ nodes.
Could someone help me to understand how to solve it and provide (if any) a usable formula for $\mu_{n}^{(m)}$?