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Q. Find the no of ways of inserting r dolllars using 1 dollar, 2 dollar and 5 dollar tokens,when order doesn't matter and when order doesn't matter. Ans. When order doesn't matter.. $(1+x+x^2+x^3+..)(1+x^2+x^4+..)(1+x^5+x^{10}+..)$ Coefficient of $x^r$ in above generating function. I clearly understand this. But I don't understand the process when order matters. Please somebody explain how (and why) to approach when order matters.

Vibhav
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2 Answers2

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Order Does Not Matter

If the order doesn't matter, then the coefficient of $x^n$ in $$ \begin{align} &\overbrace{\left(1+x+x^2+x^3+\dots\right)}^{\text{exponent = number of $\$1$}}\overbrace{\left(1+x^2+x^4+x^6+\dots\right)}^{\text{exponent = 2 $\times$ number of $\$2$}}\overbrace{\left(1+x^5+x^{10}+x^{15}+\dots\right)}^{\text{exponent = 5 $\times$ number of $\$5$}}\\ &=\bbox[5px,border:2px solid #C0A000]{\frac1{1-x}\frac1{1-x^2}\frac1{1-x^5}}\\ &\small=1+x+2x^2+2x^3+3x^4+4x^5+5x^6+6x^7+7x^8+8x^9+10x^{10}+11x^{11}+13x^{12}\\ &\small\,+14x^{13}+16x^{14}+18x^{15}+20x^{16}+22x^{17}+24x^{18}+26x^{19}+29x^{20}+\dots \end{align} $$ is the number of ways to choose ones, twos, and fives that sum to $n$.

For example, suppose $n=12$. The case of $3$ $\$1$ tokens and $2$ $\$2$ tokens and $1$ $\$5$ token is counted by the $x^3$ term from the first sum times the $x^4$ term from the second sum times the $x^5$ term from the third sum.

Since the denominator of the generating function is $1-x-x^2+x^3-x^5+x^6+x^7-x^8$, the recursion for the coefficients is $$ a_n=a_{n-1}+a_{n-2}-a_{n-3}+a_{n-5}-a_{n-6}-a_{n-7}+a_{n-8} $$

Order Matters

If the order matters, break down the generating function into the sum of the generating functions for a given number of tokens: $$ \begin{align} &1+\overbrace{\left(x+x^2+x^5\right)\!\vphantom{\left(x^2\right)^2}}^{\text{one token}}+\overbrace{\left(x+x^2+x^5\right)^2}^{\text{two tokens}}+\overbrace{\left(x+x^2+x^5\right)^3}^{\text{three tokens}}+\dots\\ &=\bbox[5px,border:2px solid #C0A000]{\frac1{1-\left(x+x^2+x^5\right)}}\\ &=1+x+2x^2+3x^3+5x^4+9x^5+15x^6+26x^7+44x^8+75x^9\\ &\,+128x^{10}+218x^{11}+372x^{12}+634x^{13}+1081x^{14}+1843x^{15}+\dots \end{align} $$ Under the six tokens term, we would have a product like $$ \small\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right)\left(x+x^2+x^5\right) $$ There is now a term to specify each possible ordering of the tokens. Using the example above, one such term for $3$ $\$1$ tokens and $2$ $\$2$ tokens and $1$ $\$5$ token would be $$ \small\overbrace{\left(x\color{#C0C0C0}{+x^2+x^5}\right)}^{\$1}\overbrace{\left(\color{#C0C0C0}{x+}x^2\color{#C0C0C0}{+x^5}\right)}^{\$2}\overbrace{\left(\color{#C0C0C0}{x+x^2+}x^5\right)}^{\$5}\overbrace{\left(x\color{#C0C0C0}{+x^2+x^5}\right)}^{\$1}\overbrace{\left(x\color{#C0C0C0}{+x^2+x^5}\right)}^{\$1}\overbrace{\left(\color{#C0C0C0}{x+}x^2\color{#C0C0C0}{+x^5}\right)}^{\$2} $$ Since the denominator of the generating function is $1-x-x^2-x^5$, the recursion for the coefficients is $$ a_n=a_{n-1}+a_{n-2}+a_{n-5} $$

robjohn
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  • I understand that. But what I don't understand is the difference in approach. In 2nd case we have generating function for each token. Then we multiplied with itself as many times as no of tokens. Why we can't use this approach in the first case, i.e. finding generating function for each token. – Vibhav Oct 20 '15 at 02:19
  • In the first approach, we account for all of the $$1$ tokens, $1+x+x^2+\dots$, before all of the $$2$ tokens, $1+x^2+x^4+\dots$, and those before all of the $$5$ tokens, $1+x^5+x^{10}+\dots$. In the second approach we have to count the tokens in different orders. Thus, we have factors of $x+x^2+x^5$ for each token. – robjohn Oct 20 '15 at 02:45
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When order does matter, the contents of the $k$-th bracket will be the possibilities for the $k$-th token to be inserted, which are $1,2$ or $5$ dollars regardless of $k$.

So a first guess for the generating function that respects order is

$$g(x)=(x+x^2+x^5)\times(x+x^2+x^5)\times(x+x^2+x^5)\times\cdots$$

but this has the problem that the degree of any term in the product is infinite.

But we know that at least $4$ and at most $19$ tokens can be used to make up the amount.

So

$$\begin{align} g(x)&=\sum_{k=4}^{19}{(x+x^2+x^5)^k} \\ &=(x+x^2+x^5)^4\cdot\sum_{k=0}^{15}{(x+x^2+x^5)^k} \\[2ex] &=(x+x^2+x^5)^4\cdot\frac{(x+x^2+x^5)^{16}-1}{x+x^2+x^5-1} \end{align}$$

Marconius
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