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?
Answer & explanation
Answer: (d) 35
99% of 400 is 396, so the student loses exactly 4 marks in all across the four papers. Sharing 4 lost marks among 4 papers (0 or more each) can be done in C(7, 3) = 35 ways.
- Total maximum = 4 × 100 = 400; 99% of 400 = 396.
- Let the marks lost in P, Q, R, S be w, x, y, z (each 0 or more) with w + x + y + z = 4; no paper can exceed 100.
- Number of solutions = C(4 + 3, 3) = C(7, 3) = 35.
- Check by cases: (4,0,0,0) type 4 ways, (3,1,0,0) 12, (2,2,0,0) 6, (2,1,1,0) 12, (1,1,1,1) 1 → 35.
Remember · Count marks lost instead of marks scored; n identical units in r boxes can be shared in C(n + r − 1, r − 1) ways.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·