A certain tree $T$ of order $n$ contains only vertices of degree $1$ and $3$. Show that $T$ contain $\frac{n-2}{2}$ vertices of degree $3$. A certain tree $T$ of order $n$ contains only vertices of degree $1$ and $3$. Show that $T$ contain $\frac{n-2}{2}$ vertices of degree $3$.
Here is what i got so far:
Since $T$ is a tree of order $n$, the size of $T$, 
$$m=n-1$$
I also know that the sum of degree of all vertices in a graph is double the size. Since $T$ contain only vertices of degree $1$ and $3$. Let $x,y$ be number of vertices of degree $1$ and $3$ respectively, we have
$$x+3y=2(n-1)$$
From here, I don't know how to get $y=\frac{n-2}{2}$
 A: Hint What is $x+y$ ?
From there you are done.
A: Here  is  an  answer  that  uses  generating  functions  and  Lagrange
inversion, to verify that it produces a correct result, to present the
method and to motivate additional exploration.

Note that  this type of  tree can be  obtained by taking  three rooted
ordered proper binary trees and attaching them to a root node. (In fact every
node of degree three in the tree can play this role.)

Now the  species $\mathcal{T}$  of binary trees  with nodes  of degree
three marked has equation
$$\mathcal{T} = \mathcal{Z} + \mathcal{U}\mathcal{Z} \mathcal{T}^2.$$
This gives the functional equation for the generating function $T(z)$
$$T(z) = z + uz T(z)^2.$$
We  extract coefficients  from this  with Lagrange  Inversion starting
from
$$[z^n] T(z) = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z) \; dz.$$
Put $T(z)=w$ so that $$z = \frac{w}{1+uw^2}$$ and
$$dz = \left(\frac{1}{1+uw^2} - \frac{w\times 2uw}{(1+uw^2)^2} \right)\; dw$$
which gives for the integral
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{(1+uw^2)^{n+1}}{w^{n+1}} 
w  \left(\frac{1}{1+uw^2} 
- \frac{2uw^2}{(1+uw^2)^2}\right)\; dw.$$
We process the two components in turn. The first is
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 
\frac{(1+uw^2)^n}{w^n} \; dw$$
and the second
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} 2u \times 
\frac{(1+uw^2)^{n-1}}{w^{n-2}} \; dw.$$
The first gives for $n$ odd the contribution 
$$u^{(n-1)/2} {n\choose (n-1)/2}$$
and the second one also for $n$ odd
$$2u \times u^{(n-3)/2} {n-1\choose (n-3)/2}.$$ 
This yields
$$u^{(n-1)/2}
\times\left({n\choose (n-1)/2} - 2{n-1\choose (n-3)/2} \right).$$
which simplifies to
$$u^{(n-1)/2} \times
\frac{1}{(n+1)/2} {n-1\choose (n-1)/2}$$
where we have encountered the Catalan numbers which are
OEIS A000108.

This shows that  every such binary tree has  $(n-1)/2$ nodes of degree
three. If we  take three such trees on $n_1, n_2$  and $n_3$ nodes and
attach  them to  a root  node the  resulting tree  has $n_1+n_2+n_3+1$
nodes and  $(n_1+n_2+n_3-3)/2+1$ nodes  of degree three  including the
root.
But we have
$$\frac{(n_1+n_2+n_3+1)-2}{2} = 
\frac{n_1+n_2+n_3-1}{2} = \frac{n_1+n_2+n_3-3}{2}+1$$
and the claim holds.
The point here is of course that this very same method can be used to prove much more difficult results including producing the distribution of the parameter marked by $u$ when it is not constant in the object space corresponding to a fixed value of the main variable.
