Correct Option (4)

The maximum marks for each of the four papers (P, Q, R, S) are 100. Therefore, the total maximum marks for the examination are 4 × 100 = 400.

To score 99% in the examination, the student must obtain 0.99 × 400 = 396 marks.

The problem requires determining the number of distinct ways (n) to achieve a total score of 396 marks across the four papers, given that marks are integers and each paper has a maximum score of 100.

This can be analyzed by considering the total "lost" marks. The difference between the maximum possible score and the target score is 400 - 396 = 4 marks. We need to distribute these 4 lost marks among the four papers (P, Q, R, S), such that each paper loses a non-negative integer amount of marks, and the total lost marks sum to 4. Each paper can lose a maximum of 4 marks (e.g., if one paper scores 96 and the others score 100).

The possible distributions of these 4 lost marks (x₁, x₂, x₃, x₄) and the corresponding number of permutations are as follows:

Case 1: (1, 1, 1, 1)

Each paper loses 1 mark, meaning each paper scores 99. The number of ways to arrange these identical lost marks is 4! / (1!1!1!1!) = 1.
Case 2: (0, 1, 1, 2)

One paper loses 0 marks (scores 100), two papers lose 1 mark each (score 99), and one paper loses 2 marks (scores 98). The number of ways to arrange these lost marks is 4! / (1!2!1!) = 24 / 2 = 12.
Case 3: (0, 0, 1, 3)

Two papers lose 0 marks each (score 100), one paper loses 1 mark (scores 99), and one paper loses 3 marks (scores 97). The number of ways to arrange these lost marks is 4! / (2!1!1!) = 24 / 2 = 12.
Case 4: (0, 0, 2, 2)

Two papers lose 0 marks each (score 100), and two papers lose 2 marks each (score 98). The number of ways to arrange these lost marks is 4! / (2!2!) = 24 / 4 = 6.
Case 5: (0, 0, 0, 4)

Three papers lose 0 marks each (score 100), and one paper loses 4 marks (scores 96). The number of ways to arrange these lost marks is 4! / (3!1!) = 24 / 6 = 4.

Summing the number of ways from all possible cases:

Total number of ways (n) = 1 + 12 + 12 + 6 + 4 = 35.

Incorrect Options:

Options 1 (16), 2 (17), and 3 (23) are incorrect. These values do not account for all possible combinations of integer marks across the four papers that sum to 396, considering the permutations of identical scores. The comprehensive enumeration of all valid distributions of lost marks and their arrangements yields a total of 35 distinct ways.