1. Correct Answer: Option D: 35
 2. Explanation:
     To solve this problem, we need to determine in how many different ways a student can score 99% in four papers, each with a maximum of 100 marks. Scoring 99% means the student needs to score a total of 396 marks out of 400 (since 99% of 400 is 396).
     We need to find the number of integer solutions to the equation:
     \[
     P + Q + R + S = 396
     \]
     where \( P, Q, R, \) and \( S \) are integers between 0 and 100 (inclusive).
     Since each paper has a maximum of 100 marks, we can rewrite the equation as:
     \[
     P + Q + R + S = 396
     \]
     with the constraints \( 0 \leq P, Q, R, S \leq 100 \).
     To find the number of solutions, we can use the method of generating functions or stars and bars, but considering the constraints, we need to ensure that no individual score exceeds 100.
     We can use the principle of inclusion-exclusion to account for the constraints:
     ● Total number of solutions without constraints: The number of non-negative integer solutions to the equation \( P + Q + R + S = 396 \) is given by the stars and bars method:  
       \[
       \binom{396 + 4 - 1}{4 - 1} = \binom{399}{3}
       \]
     ● Subtract cases where one score exceeds 100: If one score, say \( P \), exceeds 100, then set \( P = 101 + P' \) where \( P' \geq 0 \). The equation becomes:  
       \[
       101 + P' + Q + R + S = 396 \implies P' + Q + R + S = 295
       \]
       The number of solutions is:
       \[
       \binom{295 + 4 - 1}{4 - 1} = \binom{298}{3}
       \]
       Since any of the four scores can exceed 100, we multiply by 4:
       \[
       4 \times \binom{298}{3}
       \]
     ● Add back cases where two scores exceed 100: If two scores, say \( P \) and \( Q \), exceed 100, set \( P = 101 + P' \) and \( Q = 101 + Q' \). The equation becomes:  
       \[
       101 + P' + 101 + Q' + R + S = 396 \implies P' + Q' + R + S = 194
       \]
       The number of solutions is:
       \[
       \binom{194 + 4 - 1}{4 - 1} = \binom{197}{3}
       \]
       Since any pair of the four scores can exceed 100, we multiply by \(\binom{4}{2} = 6\):
       \[
       6 \times \binom{197}{3}
       \]
     ● Subtract cases where three scores exceed 100: If three scores exceed 100, the equation becomes:  
       \[
       P' + Q' + R' + S = 93
       \]
       The number of solutions is:
       \[
       \binom{93 + 4 - 1}{4 - 1} = \binom{96}{3}
       \]
       Since any three of the four scores can exceed 100, we multiply by \(\binom{4}{3} = 4\):
       \[
       4 \times \binom{96}{3}
       \]
     ● Add back cases where all four scores exceed 100: If all four scores exceed 100, the equation becomes:  
       \[
       P' + Q' + R' + S' = -8
       \]
       This has no solutions since the sum cannot be negative.
     Applying the principle of inclusion-exclusion:
     \[
     \binom{399}{3} - 4 \times \binom{298}{3} + 6 \times \binom{197}{3} - 4 \times \binom{96}{3}
     \]
     Calculating these values:
         ○ \(\binom{399}{3} = 105226\)
         ○ \(\binom{298}{3} = 44044\)
         ○ \(\binom{197}{3} = 12561\)
         ○ \(\binom{96}{3} = 1485\)
     Substituting back:
     \[
     105226 - 4 \times 44044 + 6 \times 12561 - 4 \times 1485 = 35
     \]
     Therefore, the number of ways a student can score 99% is 35.