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?
(a) 16
(b) 17
(c) 23
(d) 35
Correct Answer: (d) 35
Question from UPSC Prelims 2023 CSAT
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.
- All four papers scored 99 marks each. This is one way.
- Three papers scored 100 marks each and one paper scored 96 marks. This can happen in 4 ways (4C3) · (1C1).
- 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).
- 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).
- 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.