Dakota randomly selected three different integers 1 through 6. What is the probability that the three numbers selected could be the sides of a triangle? Express your answer as a common fraction.
+6250 \(\text{What's necessary for a triangle is that the}\\ \text{sum of the lengths of any two sides is greater than the length of the third}\\ \text{a little work shows that there are 34 valid unordered triplets that serve as triangle lengths}\\ \text{there are a total of $\dbinom{3+6-1}{6-1}=56$ total unordered triplets}\\ \text{Thus the probability is $p=\dfrac{34}{56} = \dfrac{17}{28}$}\)
.
