Question

a) Show that $1/\left(1-x-x^2-x^3-x^4-x^3-x^6\right)$is the generating function for the number of ways that the sum n can be obtained when a die is rolled repeatedly and the order of the roll matters.

b) Use part (a) to find the number of ways to roll a total of 8 when a die is rolled repeatedly, and the order of the roll matters.

Short answer

(a)The total generating function is$\frac{1}{1-x-x^2-x^3-x^4-x^5-x^6}$.

(b) The coefficient of $x^8$ is 125.

Step-by-step solution

Use Generating Function:

Generating function for the sequence${\mathit{a}}_{\mathbf{0}}\mathbf{,}{\mathit{a}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{k}}\mathbf{,}\mathbf{\dots }$of real numbers is the infinite series,

$\mathit{G}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{\mathit{a}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}\mathbf{+}\mathbf{\dots }\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}{\mathit{a}}_{\mathbf{k}}{\mathit{x}}^{\mathbf{k}}$

Extended binomial theorem;

${\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{u}}\mathbf{=}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{+}\mathbf{\infty }}\left(\begin{array}{c}u\\ k\end{array}\right){\mathit{x}}^{\mathbf{k}}$

On each roll, we have a chance to roll 1 (represented by $x$), 2 (represented by$x^2)$, 3 (represented by $x^3)$, 4 (represented by $\left.x^4\right),5\left(\right.$represented by $x^5)$and $6\left(\right.$represented by $\left.x^6\right)$.

$x+x^2+x^3+x^4+x^5+x^6$

If we then make $n$rolls, then the generating function of the $n$rolls is the product of the generating functions for each draw:

${\left(x+x^2+x^3+x^4+x^5+x^6\right)}^n$

The total generating function is then the sum over all possible rolls:

$$\begin{gathered} \sum _{n=0}^{+\infty }{\left(x+x^2+x^3+x^4+x^5+x^6\right)}^n=\frac{1}{1-\left(x+x^2+x^3+x^4+x^5+x^6\right)}\text{Since} \sum _{k=0}^{+\infty }{x}^k=\frac{1}{1-x} \\ =\frac{1}{1-x-x^2-x^3-x^4-x^5-x^6} \end{gathered}$$

Calculate all possible combinations that result in a term containing x8 in the first 8 terms of the sum.

Roll an 8 in at most 8 rolls.

When we roll 1 on each roll.

Expand the sum:

$$\begin{gathered} \sum _{k=0}^8{\left(x+x^2+x^3+x^4+x^5+x^6\right)}^k=1+x+2x^2+4x^3+8x^4+16x^5+32x^6+63x^7 \\ +125x^8+247x^9+482x^{10}+\dots \end{gathered}$$

Thus, the coefficient of $x^8$is 125.