<s>[INST] 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? 
 [/INST] 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
$$\sum_{i=1}^{18} \frac{5!}{n_i} = 651.$$
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