In how many ways can we choose one number from the set {1,2,3}, one number from the set {5,6,7}, and one number from the set {9,10,11} such that the three could be the sides of a nondegenerate triangle?

Guest Oct 15, 2020

#1**0 **

From your 3 sets, you can have this many combinations:

[(1, 2, 3), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 9), (1, 2, 10), (1, 2, 11), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 9), (1, 3, 10), (1, 3, 11), (1, 5, 6), (1, 5, 7), (1, 5, 9), (1, 5, 10), (1, 5, 11), (1, 6, 7), (1, 6, 9), (1, 6, 10), (1, 6, 11), (1, 7, 9), (1, 7, 10), (1, 7, 11), (1, 9, 10), (1, 9, 11), (1, 10, 11), (2, 3, 5), (2, 3, 6), (2, 3, 7), (2, 3, 9), (2, 3, 10), (2, 3, 11), (2, 5, 6), (2, 5, 7), (2, 5, 9), (2, 5, 10), (2, 5, 11), (2, 6, 7), (2, 6, 9), (2, 6, 10), (2, 6, 11), (2, 7, 9), (2, 7, 10), (2, 7, 11), (2, 9, 10), (2, 9, 11), (2, 10, 11), (3, 5, 6), (3, 5, 7), (3, 5, 9), (3, 5, 10), (3, 5, 11), (3, 6, 7), (3, 6, 9), (3, 6, 10), (3, 6, 11), (3, 7, 9), (3, 7, 10), (3, 7, 11), (3, 9, 10), (3, 9, 11), (3, 10, 11), (5, 6, 7), (5, 6, 9), (5, 6, 10), (5, 6, 11), (5, 7, 9), (5, 7, 10), (5, 7, 11), (5, 9, 10), (5, 9, 11), (5, 10, 11), (6, 7, 9), (6, 7, 10), (6, 7, 11), (6, 9, 10), (6, 9, 11), (6, 10, 11), (7, 9, 10), (7, 9, 11), (7, 10, 11), (9, 10, 11)], >Total distinct combinations = 84

You will have to go through each combination, add up the 2 smaller numbers and see if their total is greater than the 3rd number. Example: (1, 2, 5). 1 + 2 = 3 which smaller than 5. Therefore, it cannot make a triangle. On the other hand, (7, 9, 10) will make a triangle since 7 + 9=16 which is > 10. You will have to do this for each and every one of the above 84 combinations. Then you will have to count all the ones that will make a triangle and divide that number by 84. And that will be your Probability.

Guest Oct 15, 2020