Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$ 
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x^3+x^6+x^9+\dots$$
$$P_b(x) = 1 +x+x^2+x^3+\dots + x^9 = \frac{1-x^{10}}{1-x}$$
$$P_c(x) = 1+x^2 + x^4 + x^6 + \dots$$
I want to know if my $P_a(x)$ and $P_c(x)$ are stated correctly and how can they be simplified so that I get to $x^{15}$ in an easier way. 
Big thanks for your help. 

Um.... why do work when simply counting them is magnitudes of effort easier.  
$a = 3k; k=0... 5$.  $c$ is even so $b$ is odd if and only if $a$ and $k$ are even.  So $b$ can be any even/odd number $\le \min (15-a, 10)$ and $c = 15-a -b$.
Just count them.  If $a=0$ then $b=1,3,...,9$ and $c= 15-b$ and there are $5$ such solutions.
If $a=3$ the $b=0,...,8$ and $c = 12-b$ and there are $5$ such solutions.
If $a = 6$ then $b=1,....9$ and $c = 9-b$ and there are $5$ such solutions.
If $a = 9$ then $b=0,2,4,6$ and $c =6-b$ and there are $4$ such solutions.
If $a= 12$ then $b=1,3$ and $c=3-b$ and there are $2$ such solutions.
If $a = 15$ then $b = 0$ and $c = 0$ and there is $1$ such solution.
So there are $5+5+5+4+2 +1=22$ solutions.
