VJesus12 wrote:If three different integers are selected at random from the integers 1 through 8, what is the probability that the three selected integers can be the side lengths of a triangle?
A. 11/28
B. 27/56
C. 1/2
D. 4/7
E. 5/8
[spoiler]OA=A[/spoiler]
Source: Manhattan GMAT
In order for the three numbers selected to be the side lengths of a triangle, we need the sum of the two smallest numbers to be greater than the largest number. Thus, we have:
2 + 3 > 4, 2 + 4 > 5, 2 + 5 > 6, 2 + 6 > 7, 2 + 7 > 8 --- (5 instances if the smallest number is 2)
3 + 4 > 5, 3 + 4 > 6, 3 + 5 > 6, 3 + 5 > 7, 3 + 6 > 7, 3 + 6 > 8, 3 + 7 > 8 --- (7 instances if the smallest is 3)
4 + 5 > 6, 4 + 5 > 7, 4 + 5 > 8, 4 + 6 > 7, 4 + 6 > 8, 4 + 7 > 8 --- (6 instances if the smallest is 4)
5 + 6 > 7, 5 + 6 > 8, 5 + 7 > 8 --- (3 instances if the smallest is 5)
6 + 7 > 8 --- (1 instance if the smallest is 6)
Therefore, there are a total of 5 + 7 + 6 + 3 + 1 = 22 sets of three distinct numbers that can be selected from the integers 1 through 8 that can be the side lengths of a triangle. Since the total number of ways of selecting 3 numbers from 8 is 8C3 = (8 x 7 x 6)/(3 x 2) = 56, the probability that the 3 numbers can be the side lengths of a triangle is 22/56 = 11/28.
Answer: A
