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. 

Can't this simply be done by making use of the fact that $b + c = 15 - a$ is divisible by $3$ to conclude that $15-a$ can only equal one of $0,$ $3,$ $6,$ $9,$ $12,$ $15$? In the lists below, keep in mind that the values are nonnegative and $b = 0,\,1,\,\ldots,\,9$ and $c$ is even. Thus, for example, if $b+c = 9,$ then $c$ can only be $0,\,2,\,4,\,6,\,8$ because $c \geq 0$ is even and $b \leq 9.$ Each of these $5$ values of $c$ in this example is also easily seen correspond to a unique solution for $(a,b,c).$
$a=15$ and $b+c=0$
There is only $1$ solution, namely $(b,c) = (0,0).$
$a=12$ and $b+c=3$
There are $2$ solutions, namely $(b,c) = (1,2),\;(3,0).$
$a=9$ and $b+c=6$
There are $4$ solutions, namely $(b,c) = (0,6),\;(2,4),\;(4,2),\;(6,0).$
$a=6$ and $b+c=9$
There are $5$ solutions, namely $(b,c) = (1,8),\;(3,6),\;(5,4),\;(7,2),\,(9,0).$
$a=3$ and $b+c=12$
There are $5$ solutions, namely $(b,c) = (0,12),\;(2,10),\;(4,8),\;(6,6),\,(8,4).$ (No more, because we must have $b \leq 9.)$
$a=0$ and $b+c=15$
There are $5$ solutions, namely $(b,c) = (1,14),\;(3,12),\;(5,10),\;(7,8),\,(9,6).$
Therefore, altogether there are $1+2+4+5+5+5 = 22$ solutions.
