How many ways can 5 dice produce a total of 20? 
How many ways can $5$ dice produce a total of $20$?

I set up the equation $x_1+x_2+x_3+x_4+x_5 = 20$. The total possible number of combinations is $\binom{19}4$. From there I subtracted the number of possibilities where $1$ of the variables is greater than $6$, which I got as $5\times\binom{13}4$. I also subtracted the possibilities where $2$ variables is greater than $6$, which I used $10 \times \binom74$. I got the $10$ from the number of ways I can choose $2$ of the variables to be greater than $6$ out of the $5$ total variables.
So I have $$\binom{19}4 -5\times\binom{13}4 -10\times\binom74$$
However, I get a negative answer, which can't be right. Can anyone point a flaw in my logic? 
 A: As in this answer, we can approach this question using either generating functions or inclusion-exclusion. Instead of counting the number of ways for $5$ numbers from $1$ to $6$ to sum to $20$, we will count the number of ways for $5$ numbers from $0$ to $5$ to sum to $15$ (then add $1$ to each of the $5$ numbers).

Generating Functions
The generating function for the number of ways for $5$ integers from $0$ to $5$ to sum to a given number is
$$
\begin{align}
\hspace{-1cm}(1+x+x^2+x^3+x^4+x^5)^5
&=\left(\frac{1-x^6}{1-x}\right)^5\\
&=\sum_{k=0}^5\binom{5}{k}(-1)^kx^{6k}\sum_{j=0}^\infty\binom{-5}{j}(-x)^j\\
&=\sum_{k=0}^5\binom{5}{k}(-1)^kx^{6k}\sum_{j=0}^\infty\binom{j+4}{j}x^j\tag{1}
\end{align}
$$
We can get the coefficient of $x^{15}$ in $(1)$ by choosing $6k+j=15$:
$$
\begin{align}
\hspace{-1cm}\sum_{k=0}^5(-1)^k\binom{5}{k}\binom{15-6k+4}{15-6k}
&=\binom{5}{0}\binom{19}{15}-\binom{5}{1}\binom{13}{9}+\binom{5}{2}\binom{7}{3}\\
&=651\tag{2}
\end{align}
$$

Inclusion-Exclusion
Without restriction on the size of the terms, using the standard $\mid$ and $\circ$ argument ($15$ $\circ$s and $4$ $\mid$s), there are $\binom{15+4}{4}$ ways to choose 5 non-negative integers that sum to $15$.
$$
\text{one sum for each arrangement}\\
2+4+6+1+2=\circ\,\circ\mid \circ\circ\circ\,\circ\mid \circ\circ\circ\circ\circ\,\circ\mid\circ\mid \circ\circ
$$
Now let's count how many ways there are to have terms greater than $5$. There are $\binom{5}{1}$ ways to choose which $1$ term should be greater than $5$. To count the number of sums with $1$ term at least $6$, that would be $\binom{15-6+4}{4}$.
$$
\text{consider the red group atomic}\\
2+7+3+2+1=\circ\,\circ\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ}\circ\mid\circ\circ\circ\mid\circ\,\circ\mid\circ
$$
There are $\binom{5}{2}$ ways to choose which $2$ terms should be greater than $5$. To count the number of sums with $2$ terms at least $6$, that would be $\binom{15-12+4}{4}$.
$$
7+0+6+1+1=\color{#C00000}{\circ\circ\circ\circ\circ\circ}\circ\mid\mid\color{#C00000}{\circ\circ\circ\circ\circ\,\circ}\mid\circ\mid\circ
$$
There is no way for $3$ terms to be greater than $5$. Inclusion-Exclusion says there are
$$
\binom{19}{4}-\binom{5}{1}\binom{13}{4}+\binom{5}{2}\binom{7}{4}=651
$$
ways for $5$ terms to sum to $15$ with each term at most $5$.

Problem in the question
With Inclusion-Exclusion, the terms in the sum are alternating. If the last $-$ sign is changed to a $+$, your answer would be correct.
A: The maximum of the five dice is either $4, 5$, or $6$.  Reasoning:
If the maximum of all five is $6$, then the maximum of the remaining four is $4, 5$ or $6$, because $6+3+3+3+3 < 20.$  If the maximum of all five is $5$, then the maximum of the remaining four is $4$ or $5$, because $5+3+3+3+3 < 20.$  If the maximum of all five is $4$, then they're all $4$.  The maximum cannot be $3$ or less because $3+3+3+3+3 < 20.$
If the top two dice are $6,6$, the sum of the other three is $8$.  Five combinations with all three $\leq 6$: $$(6,1,1), (5,2,1), (4,3,1), (4,2,2), (3,3,2).$$
If the top two dice are $6,5$, the sum of the other three is $9$.  Five combinations with all three $\leq 5$: $$(5,3,1), (5,2,2), (4,4,1), (4,3,2), (3,3,3).$$
If the top two dice are $6,4$, the sum of the other three is $10$.  Two combinations with all three $\leq 4$: $$(4,4,2), (4,3,3).$$
If the top two dice are $5,5$, the sum of the other three is $10$.  Four combinations with all three $\leq 5$: $$(5,4,1), (5,3,2), (4,4,2), (4,3,3).$$
If the top two dice are $5,4$, the sum of the other three is $11$.  Just one combination with all three $\leq 4$:  $$(4,4,3).$$
Finally, there's $(4,4,4,4,4).$
So, there are $18$ combinations of $5$ dice that add up to $20$.
Edit:  If order matters, then you'll need to consider the distinct permutations of each combination above.  For each combination, there are $5!/n_i$ permutations, where
$$n_i =
  \begin{cases}
   2! = 2 & \text{if there is one pair}\\
   2!2! = 4       & \text{if there are two pair}\\
   3! = 6 & \text{if there are three of a kind}\\
   3!2! = 12 & \text{if it's a full house}\\
   5! = 120 & \text{if it's a yahtzee}\\
  \end{cases}
$$
It's straightforward to verify that
$$\sum_{i=1}^{18} \frac{5!}{n_i} = 651.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $\ds{0 < a < 1}$:
\begin{align}&\color{#c00000}{\sum_{n_{1} = 1}^{6}\ldots%
\sum_{n_{5} = 1}^{6}\ds{\delta_{\sum_{i = 1}^{5}n_{i},\ 20}}}
=\sum_{n_{1} = 1}^{6}\ldots\sum_{n_{5} = 1}^{6}
\oint_{\verts{z}\ =\ a}{1 \over z^{21 - \sum_{i = 1}^{5}n_{i}}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{21}}
\pars{\sum_{n = 1}^{6}z^{n}}^{5}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a}{1 \over z^{21}}\pars{z\,{z^{6} - 1 \over z - 1}}^{5}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}
{\pars{1 - z^{6}}^{5} \over z^{16}\pars{1 - z}^{5}}\,{\dd z \over 2\pi\ic}
=\sum_{n = 0}^{\infty}{-5 \choose n}\pars{-1}^{n}\oint_{\verts{z}\ =\ a}
{\pars{1 - z^{6}}^{5} \over z^{16 - n}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{n = 0}^{\infty}{n + 4 \choose 4}
\sum_{k = 0}^{5}{5 \choose k}\pars{-1}^{k}\ \underbrace{\oint_{\verts{z}\ =\ a}
{1 \over z^{16 - n - 6k}}\,{\dd z \over 2\pi\ic}}_{\ds{\delta_{n\ +\ 6k,\, 15}}}
\\[3mm]&=\left.\sum_{k = 0}^{5}{\bracks{15 - 6k} + 4 \choose 4}{5 \choose k}\pars{-1}^{k}\right\vert_{15\ -\ 6k\ \geq\ 0}
=\sum_{k = 0}^{2}\pars{-1}^{k}{19 - 6k \choose 4}{5 \choose k}
\end{align}

Then,
  \begin{align}&\color{#00f}{\large\sum_{n_{1} = 1}^{6}\ldots%
\sum_{n_{5} = 1}^{6}\ds{\delta_{\sum_{i = 1}^{5}n_{i},\ 20}}}
={19 \choose 4}{5 \choose 0}- {13 \choose 4}{5 \choose 1}
+{7 \choose 4}{5 \choose 2}
\\[3mm]&=3876\times 1 - 715\times 5 + 35\times 10
=\color{#00f}{\Large 651}
\end{align}

