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If three sticks are selected at random from a bag containing 9 sticks of length 2, 3, 4, 5, 6, 7, 8, 9 and 10 units, then what is the probability that a triangle can be formed with those three sticks?

 
None of these

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Triangle inequality theorem states that sum of the lengths of any two sides of a triangle must be greater than the third side.

Suppose a,b,c are the sides of a triangle. Then
a + b > c
a + c > b
b + c > a

But we don't need to check all these conditions. Just add the shorter sides and compare with the longer side.
If sum of the shorter sides is greater than the longer side, triangle can be formed. Else, triangle cannot be formed.

With this information, we can approach the given problem.

Total ways in which 3 numbers can be selected from (2, 3, 4, 5, 6, 7, 8, 9,10) = 9C3 = 84

Using the triangle inequality theorem, we can see that the following combinations cannot form a triangle.

(2,3,5), (2,3,6), (2,3,7), (2,3,8), (2,3,9), (2,3,10)
(2,4,6), (2,4,7), (2,4,8), (2,4,9), (2,4,10)
(2,5,7), (2,5,8), (2,5,9), (2,5,10)
(2,6,8), (2,6,9), (2,6,10)
(2,7,9), (2,7,10)
(2,8,10)

(3,4,7), (3,4,8), (3,4,9), (3,4,10),
(3,5,8), (3,5,9), (3,5,10),
(3,6,9), (3,6,10)
(3,7,10)

(4,5,9),(4,5,10)
(4,6,10)

i.e., 34 combinations cannot form triangle out of the total 84

P(selected sides cannot form a triangle) =34/84

P(selected sides can form a triangle) = 1−34/84=50/84=25/42

How? my ans is 11/42 please explain