3D sample space?

The minimum sum is 3:

Note that there are $3!=6$ ways of ordering 3 distinct numbers but $\frac{3!}{n!}$ ways of ordering 3 numbers with 'n' repeats

• Sum of 3: $(1,1,10) \rightarrow$ 1 way

• Sum of 4: $(1,2,1) \rightarrow$ 3 ways

• Sum of 5: $\begin{cases} 1, & (1,3,1) \rightarrow 3 ways \\ 2, & (1,2,2) \rightarrow 3 ways \end{cases} \implies$ 6 ways

• Sum of 6: $\begin{cases} 1, & (1,1,4) \rightarrow 3 ways \\ 2, & (1,2,3) \rightarrow 6 ways \\3, & (2,2,2) \rightarrow 1 way \end{cases} \implies$ 10 ways

• Sum of 7: $\begin{cases} 1, & (1,1,5) \rightarrow 3 ways \\ 2, & (1,2,4) \rightarrow 6 ways \\3, & (1,3,3) \rightarrow 3 ways \\ 4, & (2,3,2) \rightarrow 3 ways \end{cases} \implies$ 15 ways

• Sum of 8: $\begin{cases} 1, & (1,1,6) \rightarrow 3 ways \\ 2, & (1,2,5) \rightarrow 6 ways \\3, & (1,3,4) \rightarrow 6 ways \\ 4, & (2,2,4) \rightarrow 3 ways \\ 5, & (2,3,3) \rightarrow 3 ways \end{cases} \implies$ 21 ways

• Sum of 9: $\begin{cases} 1, & (1,2,6) \rightarrow 6 ways \\ 2, & (1,2,4) \rightarrow 6 ways \\3, & (1,3,3) \rightarrow 3 ways \\ 4, & (2,3,4) \rightarrow 6 ways \\ 5, & (2,5,2) \rightarrow 3 ways \\ 6, & (3,3,3) \rightarrow 1 way \end{cases} \implies$ 25 ways

$\frac{1+3+6+10+15+21+25}{6^3} \equiv \frac{81}{216} \equiv \frac{3}{8} \implies 3+8=\boxed{11}$