The minimum marks required for clearing a certain screening paper is 210 out of 300. The…
Mathematics · JEE Advanced · NTA Exams — Permutations and Combinations
The minimum marks required for clearing a certain screening paper is 210 out of 300. The screening paper consists of ‘3’ sections each of Physics, Chemistry, and Maths. Each section has 100 as maximum marks. Assuming there is no negative marking and marks obtained in each section are integers, the number of ways in which a student can qualify the examinatin is (Assuming no cut–off limit) :
- 210C3 – 90C3
- 93C3
- 213C3
- (210)3
Answer
(B) 93 C 3
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