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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.

JMoravitz
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Kishan Kumar
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    What is the coefficient of $x^{10}$ in the Taylor series at the origin of $$\frac{1}{1-x}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x^5}$$ ? – Jack D'Aurizio Oct 30 '17 at 15:59
  • I will tell you even more: due to the triple pole at $x=1$, the number of ways for changing $n$ dollars approaches $\frac{(n+1)(n+7)}{20}$ for large values of $n$. – Jack D'Aurizio Oct 30 '17 at 16:02
  • It should take about 1 minute to "brute force" through all of the solutions. – Doug M Oct 30 '17 at 16:04
  • For a typesetting note, if you want a dollar sign to appear as an actual dollar sign, put a slash in front of it like \$, otherwise it will be used to initiate in-line mathmode. – JMoravitz Oct 30 '17 at 16:04
  • Is it same as searching for coefficient of $x^{10}$ in $(1+x+x^2+x^5)^{10}$? still no improvement. – jonsno Oct 30 '17 at 16:05
  • As an additional note for writing questions and solutions, equals signs should only be used for equality, never for a "here's the next step." You wrote $10x_1+\dots=100\color{red}{=}x_1+\dots=10$, which if the equals sign in the middle were interpreted as an equals sign would imply that you wrote $100=10$ which is flat out false (in normal circumstances). Use words, or an arrow, or begin a new line in place of erroneous equals signs like that. – JMoravitz Oct 30 '17 at 16:08
  • @JackD'Aurizio IMHO it is more understandable this $$( 0,; 0,; 2),; ( 0,; 5,; 0),; ( 1,; 2,; 1),; ( 2,; 4,; 0),; ( 3,; 1,; 1),; ( 4,; 3,; 0),; ( 5,; 0,; 1),; ( 6,; 2,; 0),; ( 8,; 1,; 0),; ( 10,; 0,; 0)$$ – Raffaele Oct 30 '17 at 16:17
  • @Raffaele: I agree, but such representation does not provide any insight on the (quadratic) growth of the number of representations. – Jack D'Aurizio Oct 30 '17 at 16:29
  • @samjoe no, it is not the same as finding the coefficient of $x^{10}$ in $(1+x+x^2+x^5)^{10}$. That would have been if ten parts must be used, the order of the parts mattered and parts of size zero were allowed. E.g. a $50 followed by five $10's followed by four $0's would be considered a different outcome than three $10's followed by a $50 followed by four $0's and then the rest of the $10's. That is clearly not what is intended. – JMoravitz Oct 31 '17 at 00:41
  • @JMoravitz thanks for explanation! I am a little new to generating functions.. – jonsno Oct 31 '17 at 13:11

1 Answers1

5

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 $1$.

Markus Scheuer
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