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Here is the problem:

There are $6^3$ possible outcomes to rolling a die $3$ times. Out of these, how many yield a total of (exactly) $13$ dots?

My solution would be absolutely impractical for problems involving $4$ rolls of the die. And at higher numbers it would be downright impossible without brute-forcing it with a computer.

Is there a more elegant way to solve this type of problem?

My solution:

First we find all sets $\{a, b, c\}$ such that $a + b + c = 13; \; a, b, c \le 6$:

$\{1, 6, 6\}$

$\{2, 5, 6\}$

$\{3, 4, 6\}, \{3, 5, 5\}$

$\{4, 4, 5\}$

The total number of sets that fit these criteria is $5$. If $a \not= b \not= c$, then there exist $3!$ unique permutations of $\{a, b, c\}$. If $a = b \not= c$, then there exist $3$ unique permutations of $\{a, b, c\}$. -- There cannot be a set such that $a = b = c$.

There are $2$ sets of the first kind and $3$ of the second. It follow that the total number of triple die rolls that can fit the criteria is

\begin{equation*} 2 \cdot 3! + 3 \cdot 3 = 21 \end{equation*}


I can think of a second way to do it, which might be faster for slightly larger numbers... but essentially still comes down to brute force, not a method that can be generalized.

  • 1
    If you're going for elegant, I think the method from Fool's answer below is nice. Basically the computation is reduced to $\binom{12}{2} - 3 \cdot \binom{6}{2}$, with no brute-forcing, and no counting possibilities manually. Just some algebra.2011-09-28
  • 2
    Could whoever downvoted this explain why? And also explain why they didn't post such an explanation here?2011-09-28

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