Combinatorics - how many possible solutions are there for: $|x_1| + x_2+x_3 = 16$ How many possible solutions are there for this equation:
$|x_1| + x_2+x_3 = 16$ ; $x_1 \in Z$ $x_2,x_3 \in N$
I know it's a simple combinatorics question but I'm still having trouble figuring it out.
Since $x_1$ is an integer, there could be infinite solutions for this equation (I know this is probably not true but that is how I see it).
I know that in the case that $x_1 = 0$ there could be ${16+2-1 \choose 2-1}$ possible solutions.
The other case is when $x_1$ is non zero, when it is positive, the total # of combinations is:
${16+3-1 \choose 3-1}$ and another case where $x_1$ is negative.
Hence: ${16+2-1 \choose 2-1} + 2 \cdot {16+3-1 \choose 3-1}$
However, I checked and in my book it says that the solution is: 
${16+2-1 \choose 2-1} + 2 \cdot {15+3-1 \choose 3-1}$
I'm having trouble understanding the ${15+3-1 \choose 3-1}$ part, can anyone please explain this to me?
I also noticed that this same question was posted in this site but wasn't explained appropriately, so I posted again.
Many thanks.

If $x_1 = 0$ then there are $15$ pairs for $x_2, x_3$.
If $x_1 = \pm 1$ then there are $14$ pairs.
If $x_1 = \pm 2$ then there are $13$ pairs.
if $x_1 = \pm 13$ then there are $2$ pairs.
if $x_1 = \pm 14$ then there is $1$ pair.
So it's a stars and bars problem for the second part.  Each bin contains at least one of the $15$ object (because you're already handled the zero case) and there are three bins.  This is expressed (verbatim!) as
$${15 + 3 - 1 \choose 3-1},$$
and you get the factor of $2$ from the absolute value.
