number of non-negative integer solutions of $ax+by+cz=k$ How to find the number of non-negative integer solutions of $2x+3y+z=21$ ? 
Please Help , thanks in advance 
 A: The general problem of the title is somewhat complicated. The particular problem of the main post is straightforward. 
There are $8$ possibilities for $y$, $y=0$ to $y=7$.
If $y=0$, we need to make up $21$ dollars in $2$ dollar coins and $1$ dollar coins. The number of $2$ dollar coins is $0$ to $10$, so there are $11$ possibilities.
If $y=1$, we want to make up $18$ dollars. The number of $2$ dollar coins is $0$ to $9$, so there are $10$ possibilities.
If $y=2$, the same reasoning gives $8$ possibilities. 
If $y=3$ we get $7$ possibilities. 
For $y=4$ and $y=5$, we get respectively $5$ possibilities and $4$. 
For $y=6$ and $y=7$, we get respectively $2$ possibilities and $1$. 
Add up.
Remark: For the specific problem, what is needed is counting that is organized enough to ensure that we do not miss anything, and that we do not double-count.
We can use the same basic reasoning to count the number of solutions of $2x+3y+z=6k$, $6k+1$, $6k+2$, $6k+3$, $6k+4$, and $6k+5$.  For each of these we will get an explicit formula in terms of $k$. 
A: Since $x$ , $z$ are non negative integers, possible Value of $y$  are $0,1,2,3,4,5,6,7$ Now take each value of $y$ and solve the diphantine equation. For example
Taking $y=0$,,
$2x+z=21$.  So for every odd less then or equal to 21  will give a solution.
A: Since 21 is a small number, making cases will do.
However, for large numbers use the multinomial theorem.
A: We can look at the number of $(x,y)$ pairs such that $2x+3y\le 21$.
We have:
$$(0,0),(0,1),\dots,(0,7)$$
$$(1,0),\dots,(1,6)$$
$$(2,0),\dots,(2,5)$$
$$\vdots$$
$$(10,0)$$
The maximum $y$ takes the value $\lfloor \frac{21-2x}{3} \rfloor$, i.e.
$$\begin{array} {c|c}
x&\lfloor \frac{21-2x}{3} \rfloor\\
\hline
0&7\\
1&6\\
2&5\\
3&5\\
4&4\\
5&3\\
6&3\\
7&2\\
8&1\\
9&1\\
10&0\\
\hline
\sum&37
\end{array}
$$
The lists are zero-based, so the final answer is $37+11=48$.
