How many integer solutions are there to this equation? How many nonnegative integer solutions are there to: $x_1 + 2x_2 + x_3 = 10$? I am able to do this when it is simply $x_1 + x_2 + x_3 = 10$ but by multiplying $x_2$ by 2 has confused me.
 A: Like mentioned in the comments, this sort of problems must be solved pretty much case by case, but let's try to solve at least a little bit more general problem than the particular one asked, namely how many non-negative integer solutions does the equation
$$x_1 + tx_2 + x_3 = n$$
have, where $t, n\in\mathbb{N}$.
This can be done by letting $x_1$ run through $1, 2, \dots, n$ and checking how many values the variable $x_2$ can take, so that $x_1+tx_2 \leq n$. The third variable $x_3$ will then always be forced and the solution will work (this happens because the coefficient of $x_3$ is $1$, otherwise $n-x_1-tx_2$ would need to be a multiple of the coefficient of $x_3$ to get an integer solution).
Ok, so if $x_1 = j$, for $n-j-tx_2$ to remain nonnegative, we must have
$$x_2 \leq \frac{n-j}{t}$$
and $x_2$ can take the values $0, 1, \dots, \lfloor \frac{n-j}{t} \rfloor$. Therefore the number of solutions is
$$\sum_{j=0}^n \left( 1 + \left\lfloor \frac{n-j}{t} \right\rfloor \right)
= \sum_{j=0}^n \left( 1 + \left\lfloor \frac{j}{t} \right\rfloor \right)
$$
$$
=n+1 + \sum_{j=0}^n \left\lfloor \frac{j}{t} \right\rfloor
$$
(Above we just flipped the sum, i.e change of index $j\to n-j$ for a nicer looking formula).
For the particular case $t=2$, we can simplify this by considering cases $n$ even and $n$ odd, splitting the sum to even and odd parts and using the fact that $\sum_{k=0}^r k= \frac{r(r+1)}{2}$ to
$$n+1 + \left \lfloor \frac{n^2}{4} \right \rfloor$$
and for $n=10$ we get $36$ solutions.
A: $x_2$ may be $0...5$.  $x_1$ may be $0.. ,10-2x_2$ and $x_3$ is fixed at $10-2x_2 - x_1$
So there are
$\sum\limits_{x_2=0}^5 \sum\limits_{x_1=0}^{10-2x_2} 1$
Which may be reindexed as 
$\sum\limits_{x_2=0}^5 \sum\limits_{x_1=10; -1}^{2x_2} 1=$
$\sum\limits_{x_2=0}^5 \sum\limits_{x_1=0}^{2x_2} 1=$
$\sum\limits_{x_2=0}^5 (2x+1) = $
$6^2 = 36$
A: Perhaps the hope for any general methods was too hastily discarded. We can use generating functions.
If $a_n$ denotes the number of solutions to
$$c_1x_1 + c_2x_2+\dots + c_kx_k = n$$
$$x_j \geq 0$$
Then the generating function $A(z) = \sum_{n=0}^{\infty} a_nz^n$ is given by
$$A(z) = \frac{1}{(1-z^{c_1})(1-z^{c_2})\dots(1-z^{c_k})}$$
This comes from the fact that compositions of $n$ into $k$ parts corresponds to a integer-sequence of length $k$. Now, you can use a size-function $x\mapsto c_jx$ for each class of integers $\cal{I}_j$ in the product $\cal{I}_1 \times \cal{I}_2 \dots \times \cal{I}_k$, so you have the correct equation. The generating function of the class $\cal{I}_j$ is $\frac{1}{1-z^{c_j}}$ because ${\cal{I}}_j = \text{SEQ}(\{ Z \})$ and the element $Z$ has size $c_j$.
I don't know if this actually helps, maybe you can do partial fraction expansion of $A(z)$.
By the way, notice how in the case when all $c_j=1$ this amounts to the usual compositions:
$$A(z) = \left( \frac{1}{1-z} \right)^k = (1-z)^{-k} = \sum_{j=0}^\infty {{-k}\choose{j}} (-z)^{j}$$
so $a_j = {{-k}\choose{j}} (-1)^{j} = {{k+j-1}\choose{k-1}}$.
