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 ✓ UPSC's answer
Why the answer is (d)
• The total maximum marks for the four papers P, Q, R, and S is 400.
• Scoring 99% means the student must obtain exactly 396 marks (0.99 * 400).
• This is equivalent to losing exactly 4 marks from the maximum possible score (400 - 396 = 4).
• Let the marks lost in papers P, Q, R, and S be p, q, r, and s respectively. Since marks are integers and cannot be negative, p, q, r, s must be non-negative integers.
• The problem reduces to finding the number of non-negative integer solutions to the equation p + q + r + s = 4.
• Using the stars and bars formula, the number of solutions is C(n + k - 1, k - 1), where n=4 (marks lost) and k=4 (papers). Thus, n = C(4 + 4 - 1, 4 - 1) = C(7, 3) = 35.
Why the other options are wrong
- (a) 16
- 16 is incorrect because it does not account for all combinations of distributing the 4 lost marks among the 4 papers.
- (b) 17
- 17 is incorrect because it fails to apply the correct combinatorial formula for distributing indistinguishable items into distinct bins.
- (c) 23
- 23 is incorrect because it is not the result of the stars and bars calculation C(7,3).
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2023, held on 28 May 2023. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.