# Find the number of ways to obtain a total of $15$ points by throwing $4$ different dice?

I've been assigned this problem:

Find the number of ways to obtain a total of $15$ points by throwing $4$ different dice?

The problem can be stated as:

$$x_1+x_2+x_3+x_4=15\text{ with } 1\leq x_n \leq 6$$

And also as coefficient of $x^{15}$ in the generating function:

$$\left[ \sum_{r=1}^6x^r \right]^4$$

But I don't know how to compute the coefficient of $x^{15}$, until now, most problems involved the product of small generating functions and it was easy to calculate by summing the powers up to a certain number. But in this case, it's not. What could be done to calculate this by hand? I have the slight impression that I could write:

$$\left[ \sum_{r=1}^6x^r \right]^4=\left[\frac{1-x^{n+1}}{1-x}\right]^4$$

But I don't know what to do from here (considering its possible/meaninful to do this).

We have

$$\sum_{r=1}^6 x^r=\frac{x-x^7}{1-x}$$

Thus using binomial theorem we have

\begin{aligned}&\left[\sum_{r=1}^6 x^r\right]^4=x^4(1-x^6)^4(1-x)^{-4}\\ =&x^4\left(\sum_{r=0}^4(-1)^r\binom{4}{r}x^{6r}\right)\left(\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\right)\\ =&\left(\sum_{r=0}^4(-1)^r\binom{4}{r}x^{6r+4}\right)\left(\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\right)\\ =&\sum_{r=0}^4\left[(-1)^r\binom{4}{r}x^{6r+4}\left(\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\right)\right]\\ =&\sum_{r=0}^4\left(\sum_{k=0}^\infty\left((-1)^r\binom{4}{r}x^{6r+4}\right)\left((-1)^k\binom{-4}{k}x^k\right)\right)\\ =&\sum_{r=0}^4\sum_{k=0}^\infty\left((-1)^{r+k}\binom{4}{r}\binom{-4}{k}x^{k+6r+4}\right)(*) \end{aligned}

Here we recall the definition of binomial coefficient: for any $\alpha\in\mathbb{R}$ and $k\in\mathbb{N}$, the binomial coefficient "$\alpha$ choose $k$" is

$$\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\dots(\alpha-(k-1))}{k!}$$

and the binomial theorem says

$$(a+b)^{\alpha}=\sum_{k=0}^\infty\binom{\alpha}{k}a^{\alpha-k}b^k$$

Now we look at the $x^{15}$ term in $(*)$, this is equivalent to solve $k+6r+4=15$ where $r=0,1,2,3,4$ and $k=0,1,2,\dots$. The only possibilities are $r=0,k=11$ and $r=1, k=5$, and the correponding coefficients are $(-1)^{11}\binom{4}{0}\binom{-4}{11}$ and $(-1)^{6}\binom{4}{1}\binom{-4}{5}$, by adding them we get the coefficient for the $x^{15}$ term :

\begin{aligned}&(-1)^{11}\binom{4}{0}\binom{-4}{11}+(-1)^6\binom{4}{1}\binom{-4}{5}\\ =&(-1)^{11}\frac{(-4)(-5)\dots(-4-(11-1))}{11!}+4\cdot(-1)^6\frac{(-4)(-5)\dots(-4-(5-1))}{5!}\\ =&\frac{4\cdot 5\cdot\dots\cdot 14}{11!}-4\frac{4\cdot\dots\cdot 8}{5!}\\ =&\frac{14!}{3!11!}-4\frac{8!}{3!5!}\\ =&140 \end{aligned}

• Great! Thank you very much. – Billy Rubina Feb 8 '15 at 4:11
• Excuse me, isn't it: $\displaystyle \sum_{r=1}^6 x^r=\frac{1-x^7}{1-x}$ ? – Billy Rubina Feb 8 '15 at 4:38
• If sum from $r=0$, then it is $\sum_{r=0}^6x^r=\frac{1-x^7}{1-x}$ – Frank Lu Feb 8 '15 at 4:41
• Why if the sum from $r=0$ make it become $\frac{x-x^7}{1-x}$? – Billy Rubina Feb 8 '15 at 4:44
• perhaps this is helpful:en.wikipedia.org/wiki/Geometric_series – Frank Lu Feb 8 '15 at 4:46

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Each argument $\ds{\delta_{x_{1} + \cdots + x_{4}\,,15}\ }$ in the multiple sum $\ds{\sum_{x_{1}=1}^{6}\ldots\sum_{x_{4}=1}^{6} \delta_{x_{1} + \cdots + x_{4}\,,15}\ }$ is equal to $\underline{one}$ whenever $\ds{x_{1} + \cdots + x_{4}=15}$ and it becomes $\underline{zero}$ otherwise such that the sum counts the number of combinations of $\ds{x_{j}, j=1,2,3,4}$ which add to $\underline{15}$.

$\ds{\delta_{m,n}}$ is the Kronecker Delta: It's equal to $\underline{one}$ when $\ds{m=n}$ and it's equal to $\underline{zero}$ when $\ds{m \not= n}$. In order to perform the multiple sum it's convenient to use a Kronecker Delta integral representation: $$\delta_{m,n}=\oint_{\verts{z}=R}\ \frac{1}{z^{-m + n + 1}}\,\frac{\dd z}{2\pi\ic}\,;\qquad R > 0\,, \quad m,n\in{\mathbb Z}$$

We'll choose a value of $\ds{R}$ that satisfies $\ds{0 < R < 1}$: It lets us to enclose just one pole in a complex integration $\ds{\pars{~\mbox{see below}~}}$:

\begin{align}&\color{#66f}{% \sum_{x_{1}=1}^{6}\ldots\sum_{x_{4}=1}^{6}\delta_{x_{1} + \cdots + x_{4}\,,15}}\ =\sum_{x_{1}=1}^{6}\ldots\sum_{x_{4}=1}^{6}\oint_{\verts{z}=R}\ \frac{1}{z^{-x_{1} - \cdots - x_{4} + 16}}\,\frac{\dd z}{2\pi\ic} \\[5mm]&=\oint_{\verts{z}=R}\frac{1}{z^{16}}\sum_{x_{1}=1}^{6}z^{x_{1}} \sum_{x_{2}=1}^{6}z^{x_{2}}\sum_{x_{3}=}^{6}z^{x_{3}}\sum_{x_{4}=1}^{6}z^{x_{4}} \,\frac{\dd z}{2\pi\ic} \end{align} As all the sums over $\ds{x_{j},j=1,2,3,4}$, are identical, the last expression leads to: \begin{align} &\color{#66f}{% \sum_{x_{1}=1}^{6}\ldots\sum_{x_{4}=1}^{6}\delta_{x_{1} + \cdots + x_{4}\,,15}}\ =\oint_{\verts{z}=R} \frac{1}{z^{16}}\pars{\sum_{x=1}^{6}z^{x}}^{4}\,\frac{\dd z}{2\pi\ic} \\[5mm]&=\oint_{\verts{z}=R} \frac{1}{z^{16}}\pars{z\,\frac{1 - z^{6}}{1 - z}}^{4}\,\frac{\dd z}{2\pi\ic} =\oint_{\verts{z}=R} \frac{\pars{1 - z^{6}}^{4}}{z^{12}\,\pars{1 - z}^{4}}\,\frac{\dd z}{2\pi\ic} \end{align}

In order to perform the complex integration we'll use the expansion $\ds{\frac{1}{\pars{1 - z}^{4}}=\sum_{j=0}^{\infty}\ \overbrace{\pars{-4 \atop \phantom{-}j}} ^{\dsc{\pars{j + 3 \atop 3}\pars{-1}^{j}}}\pars{-z^{j}} =\sum_{j=0}^{\infty}\pars{j + 3 \atop 3}z^{j}}$ and the $\ds{\pars{1 - z^{6}}^{4}}$ binomial expansion:
\begin{align} &\color{#66f}{% \sum_{x_{1}=1}^{6}\ldots\sum_{x_{4}=1}^{6}\delta_{x_{1} + \cdots + x_{4}\,,15}}\ =\sum_{k=0}^{4}\binom{4}{k}\pars{-1}^{k} \sum_{j=0}^{\infty}\binom{j + 3}{j}\ \overbrace{\oint_{\verts{z}=R} \frac{1}{z^{12 - 6k - j}}\,\frac{\dd z}{2\pi\ic}} ^{\dsc{\delta_{6k + j,11}}} \\[5mm]&=\left.\sum_{k=0}^{4}\sum_{j=0}^{\infty}\binom{4}{k}\binom{j + 3}{3}\pars{-1}^{k} \right\vert_{\, 6k + j=11} \end{align} The values of $\ds{k,j}$ which satisfy $\ds{6k + j = 11}$ are $\ds{\pars{k=0,j=11}}$ and $\ds{\pars{k=1,j=5}}$ such that: \begin{align} &\color{#66f}{\large% \sum_{x_{1}=1}^{6}\ldots\sum_{x_{4}=1}^{6}\delta_{x_{1} + \cdots + x_{4}\,,15}}\ =\binom{4}{0}\binom{14}{3}\pars{-1}^{0} +\binom{4}{1}\binom{8}{3}\pars{-1}^{1} \\[5mm]&=1\times 364 - 4\times 56=\color{#66f}{\large 140} \end{align}

• Thanks for your answer. Could you clarify a little bit on your steps? Your answer seems really interesting, but it's not really clear what you are doing there: It seems that you're using techniques of analysis (some kind of treatment of series I never seen before). [Please, do not erase your answer because of my stupidity, leave it there and write a little bit on every step. Some day this answer might enhance my knowledge a lot]. – Billy Rubina Feb 10 '15 at 5:09
• @Vÿska Thanks. I'll do it: I'll include several details. Be patient. – Felix Marin Feb 10 '15 at 19:11
• Who downvoted this?! – Billy Rubina Feb 13 '15 at 6:48
• @Vÿska I add some details to the answer. Don't worry for down-votes. It's part of the business. I have "my" 'bullying hunter' which always check my answers, my comments and even my $\LaTeX$ edition. I don't care. Instead of learning something new he/she prefers to show his/her jealous behavior. – Felix Marin Feb 13 '15 at 20:07