NTA Abhyas JEE Main2020MathematicsPermutation and CombinationPractice
If the total number of positive integral solutions of 15 < x 1 + x 2 + x 3 ≤ 20 is k , then the value of k 100 is equal to
Correct answer
6.85
Step-by-step solution
15 < x 1 + x 2 + x 3 ≤ 2 0 ⇒ x 1 + x 2 + x 3 = 16 + r , where r = 0 ,1 , 2 ,3 , 4 . Now, the number of positive integral solutions of x 1 + x 2 + x 3 = 16 + r is 16 + r - 1 C 3 - 1 = 15 + r C 2 Thus, the required number of solutions is ∑ r = 0 4 15 + r C 2 = 15 C 2 + 16 C 2 + 17 C 2 + 18 C 2 + 19 C 2 = 15 C 2 - 16 C 3 + 16 C 3 + 16 C 2 + 17 C 2 + 18 C 2 + 19 C 2 = 15 C 2 - 16 C 3 + 20 C 3 = 68 5