Q. In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n?

Explanation : 

The total marks for the four papers is 400 (100 marks each for P, Q, R and S).

If a student scores 99% in the examination, then the total marks scored by the student is 99% of 400 = 396 marks.

Now, we need to find out the number of ways in which 396 can be expressed as a sum of four integers, each ranging from 0 to 100.

The problem can be solved by considering all possible combinations and permutations of the scores in the four papers.

1. All four papers scored 99 marks each. This is one way.
2. Three papers scored 100 marks each and one paper scored 96 marks. This can happen in 4 ways (⁴C₃) · (¹C₁).
3. Two papers scored 100 marks each, one paper scored 98 marks and one paper scored 98 marks. This can happen in 6 ways (⁴C₂) · (²C₂).
4. Two papers scored 100 marks each, one paper scored 99 marks and one paper scored 97 marks. This can happen in 12 ways (⁴C₂ · ²C₁ · ¹C₁).
5. One paper scored 100 marks, two papers scored 99 marks each and one paper scored 98 marks. This can happen in 12 ways (⁴C₁ · ³C₂ · ¹C₁).

So, the total number of ways is 1 + 4 + 6 + 12 + 12 = 35.

Hence, the correct answer is (d) 35.