Rolling three dice - sum of two results equals third one Is my answer correct?

Three six-sided dice are thrown, what is the probability that the sum of two of them equals the third one?

If order does not matter, there are 9 such possibilities:


*

*(1, 1, 2)

*(1, 2, 3)

*(1, 3, 4)

*(1, 4, 5)

*(1, 5, 6)

*(2, 2, 4)

*(2, 3, 5)

*(2, 4, 6)

*(3, 3, 6)


The total number of possibilities is $\frac{6^3}{3!}=6^2=36$. Therefore, the probability is $\frac{9}{36}$.

If you imagine there are three distinguishable dices (either different colors, or you throw them one after the other), then there are $6^3=216$ possible cases.
Now your enumeration counts down to :
$(1,1,2)$ : $3$ cases
$(1, 2, 3)$ : $6$ cases
$(1, 3, 4)$ : $6$ cases
$(1, 4, 5)$ : $6$ cases
$(1, 5, 6)$ : $6$ cases
$(2, 2, 4)$ : $3$ cases
$(2, 3, 5)$ : $6$ cases
$(2, 4, 6)$ : $6$ cases
$(3, 3, 6)$ : $3$ cases
for a total of $45$ cases, and a probability of $\frac{45}{216}=\frac{5}{24}$.
Hope I did it the right way :-(
