Find number of positive integer solutions to the equation $a+b+c=15$ where a,b are even and c is odd This is equivalent to:
Coefficient of $x^{15}$ in $(1+x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
Or
Coefficient of $x^{14}$ in $(1+x^2+x^4+\cdots+x^{14})^3$
Is this approach correct?
 A: It's already been pointed out that the OP has an error in that $a,b,c$ are supposed to be positive, not non-negative.
But I would like to point out that the algebra is a little simpler if we use  the constraints $a,b,c \ge 1$ without taking into account our knowledge that $a,b,c \le 15$.  This gives the generating function
$$\begin{align} 
f(x) &= (x^2 + x^4 + x^6 + \dots)^2 (x + x^3 + x^6 + \dots) \\
&= x^5 (1 + x^2 + x^4 + \dots)^3 \\
&= x^5 \left( \frac {1}{1-x^2} \right)^3 \\
&= x^5 (1-x^2)^{-3} \\
&= x^5 \sum_{i=0}^{\infty} \binom{3+i-1}{i} x^{2i}
\end{align}$$
From the last equation we see that the coefficient of $x^{15}$ is $\binom{7}{5} = 21$.
A: Why not just solving the problem with a direct approach? You know that $a$ and $b$ are even and $c$ is odd, so $a = 2A, b = 2B$, and $c = 2C+1$, with $A,B$ positive integers and $C \geq 0$. Now the problem amounts to solving
$$2(A+B+C)+1 = 15,$$
so you want $A,B > 0$ and $C \geq 0$ integers such that
$$A+B+C = 7.$$
You can just do this by trying case by case.
A: You say "positive" rather than "non-negative", so it is

*

*the coefficient of $x^{15}$ in $(x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
which is

*

*the coefficient of $x^{15}$ in $(x^2+x^4+\cdots+x^{12})^2(x+x^3+\cdots+x^{11})$

*the coefficient of $x^{16}$ in $(x^2+x^4+\cdots+x^{12})^3$

*the coefficient of $x^{10}$ in $(1+x^2+x^4+\cdots+x^{10})^3$

*the coefficient of $x^{5}$ in $(1+x+x^2+\cdots+x^{5})^3$
which is ${7 \choose 2}=21$
That is short enough to list as a check:
2 + 2 + 11
2 + 4 + 9
2 + 6 + 7
2 + 8 + 5
2 + 10 + 3
2 + 12 + 1
4 + 2 + 9
4 + 4 + 7
4 + 6 + 5
4 + 8 + 3
4 + 10 + 1
6 + 2 + 7
6 + 4 + 5
6 + 6 + 3
6 + 8 + 1
8 + 2 + 5
8 + 4 + 3
8 + 6 + 1
10 + 2 + 3
10 + 4 + 1
12 + 2 + 1

