# 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 (
^{4}C_{3}) · (^{1}C_{1}). - Two papers scored 100 marks each, one paper scored 98 marks and one paper scored 98 marks. This can happen in 6 ways (
^{4}C_{2}) · (^{2}C_{2}). - Two papers scored 100 marks each, one paper scored 99 marks and one paper scored 97 marks. This can happen in 12 ways (
^{4}C_{2}·^{2}C_{1}·^{1}C_{1}). - One paper scored 100 marks, two papers scored 99 marks each and one paper scored 98 marks. This can happen in 12 ways (
^{4}C_{1}·^{3}C_{2}·^{1}C_{1}).

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

Hence, the correct answer is (d) 35.