Q55. 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?

The total maximum marks for the four papers (P, Q, R, S) is 4 * 100 = 400. A student scores 99% of the total marks, which is 0.99 * 400 = 396 marks. This means the student lost a total of 400 - 396 = 4 marks across the four papers.

Let p, q, r, and s be the marks lost in papers P, Q, R, and S, respectively. Since marks scored are integers, marks lost must also be non-negative integers. The sum of marks lost is p + q + r + s = 4.

Also, since each paper has a maximum of 100 marks, the marks lost in any single paper cannot exceed 4 (e.g., if 5 marks were lost in one paper, the score would be 95, and even if other papers were 100, the total would be 395, less than 396). This problem is a classic 'stars and bars' problem, finding the number of non-negative integer solutions to x1 + x2 + ...

+ xk = n. Here, n=4 (total marks lost) and k=4 (number of papers). The number of ways is C(n + k - 1, k - 1) = C(4 + 4 - 1, 4 - 1) = C(7, 3). C(7, 3) = (7 × 6 × 5) / (3 × 2 × 1) = 35.

This question tests combinatorics, specifically the stars and bars method, a common advanced topic in CSAT.