Number of ways to score 99% across 4 papers
Question
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 ?
Options
16
17
23
35
Explanation
The maximum marks for 4 papers combined = 4 × 100 = 400. To score exactly 99\%, the total marks obtained must be: 0.99 × 400 = 396.
This means the student can lose a total of exactly 400 - 396 = 4 marks across the four papers. Let the marks lost in papers P, Q, R, and S be x_1, x_2, x_3, x_4. We must solve the non-negative integer equation: x_1 + x_2 + x_3 + x_4 = 4
Using the classic Stars and Bars formula n + r - 1{r - 1}, where n = 4 (marks lost) and r = 4 (papers): Ways = 4 + 4 - 1{4 - 1} = 7{3} = 7 × 6 × 5/3 × 2 × 1 = 35.
Since each individual loss cannot exceed 4, no paper violates the individual cap of 100 marks. Thus, n = 35.
Answer: (d).
Question details
Year
2023
Paper
CSAT
Question
Q65
Section
Numerical Ability
Sub-topic
Permutations & Combinations
Type
Number theory
Difficulty
Hard
Source hint
Number theory
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