Find the generating function for the number of solutions (a) Find the generating function for the number of solutions to
$x_1+x_2+x_3+x_4+x_5+x_6=n$ where $x_1,x_2,x_3$ are even $x_4,x_5,x_6$ are odd 
(b) Find the generating function for the number of solutions  to
$x_1+x_2+x_3+x_4+x_5=n$
subject to the condition that $0\leq x_i \leq 12$. answer in the closed form..
As i know that the number solution for the equation $x_1+x_2+x_3+x_4+...+x_k=n$ is C(n+k-1 , k)

Here is a slightly different approach.   

Ad Problem  (a):
  
  
*
  
*Since even  $x_j, 1\leq j\leq 3$ may have values $0,2,4,\ldots$ we encode candidates for even $x_j$ via $t^0+t^2+t^4+\cdots$
  
*Since odd $x_j, 4\leq j\leq 6$ may have values $1,3,5,\ldots$ we encode candidates for odd $x_j$ via $t^1+t^3+t^5+\cdots$
The contribution  of even  $x_1,x_2,x_3$ is
  \begin{align*}
 (1+t^2+t^4+\cdots)^3=\frac{1}{\left(1-t^2\right)^3}
 \end{align*}
The contribution of odd $x_4,x_5,x_6$is
  \begin{align*}
 (t^1+t^3+t^5+\cdots)^3=t^3(1+t^2+t^4+\cdots)^3=\frac{t^3}{\left(1-t^2\right)^3}
 \end{align*}
Here we apply the geometric series expansion.

It is convenient to use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ of a series.

We conclude: A generating  function $A(t)$ providing with $[t^n]A(t)$ the number of wanted solutions is
   \begin{align*}
 A(t)=\frac{t^3}{\left(1-t^2\right)^6}
 \end{align*}

$$ $$

Ad Problem  (b):
  
  
*
  
*Since  $x_j,1\leq j\leq 5$ may have values $0\leq x_j\leq 12$ we encode the contribution of $x_j$ via $t^0+t^1+t^2+\cdots+t^{12}$.
  
  
  So,  the contribution of  each  $x_j$ can be encoded as
   \begin{align*}
 1+t^1+t^2+\cdots+t^{12}=\frac{1-t^{13}}{1-t}
 \end{align*}
Here we apply the formula for the finite geometric series.
We conclude: A generating function $B(t)$ providing with $[t^n]B(t)$ the number of wanted solutions is
   \begin{align*}
 B(t)=\left(\frac{1-t^{13}}{1-t}\right)^5
 \end{align*}

Hint: If additionally the number of solutions, i.e. the coefficient of the generating functions $A(t)$, resp. $B(t)$ is to extract, it is again convenient to apply the coefficient of operator as well as the Iverson bracket.
