How to find the number of solutions for $x_1+2x_2+5x_3=10$? This is a question taken from Discrete mathematics by Kenneth Rosen:
Find the number of ways to make change for \$100 using \$10, \$20 and \$50 bills.
My approach:
Let number of \$10 notes be $x_1$.
Let number of \$20 notes be $x_2$.
Let number of \$50 notes be $x_3$.
Then the number of ways is equal to
number of solutions of $10x_1+20x_2+50x_3=100$ and so $x_1+2x_2+5x_3=10$.
Now I don't know how to find the number of solutions to this equation.


first variant
When looking for non-negative integral solutions $x_1,x_2,x_3$ we notice that the possible solutions of $x_3$ in
  \begin{align*}
  x_1+2x_2+5x_3=10\tag{1}
  \end{align*}
are $x_3\in\{0,1,2\}$ since $0\leq 5x_3\leq 10$.
Setting these three values for $x_3$ we consider instead of (1) the three equations
  \begin{align*}
  x_1+2x_2&=10\\
  x_1+2x_2&=5\\
  x_1+2x_2&=0\\
\end{align*}
The first equation has $6$ admissible values $x_2\in\{0,1,2,3,4,5\}$, the second equation has $3$ admissible values $x_2\in\{0,1,2\}$ and the third equation has one admissible value $x_2\in\{0\}$. The value of $x_1$ is then uniquely determined.
The number of admissible solutions of (1) is
  \begin{align*}
\color{blue}{6+3+1=10}
\end{align*}

second variant
We follow the comment from @JackDAurizio and use generating functions to find the number of admissible solutions.
Values of $x_3$ represent zero or more multiples of $5$ which can be encoded as
\begin{align*}
  1+x^5+x^{10}+\cdots=\frac{1}{1-x^5}
  \end{align*}
We argue similarly when considerung values of $x_1$ and $x_2$. Since the right hand side of (1) is $10$ we look for the coefficient of $x^{10}$ in
\begin{align*}
  \frac{1}{1-x^5}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}
  \end{align*}
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series $A(x)$. This way we can write e.g.
\begin{align*}
[x^n]A(x)=[x^n]\sum_{j=0}^\infty a_jx^j=a_n
\end{align*}

We obtain
  \begin{align*}
\color{blue}{[x^{10}]}&\color{blue}{\frac{1}{1-x^5}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}}\\
&=[x^{10}](1+x^5+x^{10})\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}\tag{2}\\
&=\left([x^{10}]+[x^5]+[x^0]\right)\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x}\tag{3}\\
&=[x^{10}](1+x^2+x^4+x^6+x^8+x^{10})\cdot\frac{1}{1-x}\\
&\qquad +[x^5](1+x^2+x^4)\frac{1}{1-x}+[x^0]\frac{1}{1-x}\tag{4}\\
&=6+3+1\tag{5}\\
&\color{blue}{=10}
\end{align*}
  showing the number of solutions is $10$.

Comment:


*

*In (2) we expand $\frac{1}{1-x^5}$ up to $x^{10}$ since other terms do not contribute to $[x^{10}]$.

*In (3) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

*In (4) we expand $\frac{1}{1-x^2}$ similarly as we did in (2).

*In (5) we notice that we could work as we did in (3) and since $\frac{1}{1-x}=1+x+x^2+\cdots$ each term has a contribution of $1$.
