Dakota randomly selected three different integers 1 through 8. What is the probability that the three numbers selected could be the sides of a triangle? Express your answer as a common fraction.
8 C 3 ==56 combinations
(2, 3, 4) , (2, 4, 5) , (2, 5, 6) , (2, 6, 7) , (2, 7, 8) , (3, 4, 5) , (3, 4, 6) , (3, 5, 6) , (3, 5, 7) , (3, 6, 7) , (3, 6, 8) , (3, 7, 8) , (4, 5, 6) , (4, 5, 7) , (4, 5, 8) , (4, 6, 7) , (4, 6, 8) , (4, 7, 8) , (5, 6, 7) , (5, 6, 8) , (5, 7, 8) , (6, 7, 8) , Total = 22 such triplets
The probability is: 22 / 56 == 11 / 28