Question

In a paper there are 3 sections A, B and C which have 8, 6 and 6 questions each. A student have to attempt 15 questions such that they have to attempt at least 4 questions out of each sections. Then number of ways of attempting these questions are

• A. 11,300
• B. 11,376
• C. 12,576
• D. 13,372

Answer 1

The Correct Option is B

To solve the problem of determining the number of ways to attempt 15 questions from three sections A, B, and C, where each section contains 8, 6, and 6 questions respectively, and where the student must attempt at least 4 questions from each section, follow these steps: 

1. Determine the minimum requirement:
   • Section A: At least 4 questions
   • Section B: At least 4 questions
   • Section C: At least 4 questions
2. Find the remaining number of questions:
   • Total questions to attempt = 15
   • Minimum required by condition = 12
   • Remaining questions to choose freely = \(15 - 12 = 3\)
3. Apply combinations to find the number of ways for distribution:

This problem can be viewed as a problem of distributing 3 identical items (remaining questions) into 3 distinct boxes (sections) with each box having some limits (as each section has limited number of questions). This is a classic problem of "stars and bars" in combinatorics.

• Using the formula for stars and bars, number of ways to distribute \(n\) identical items into \(r\) groups is given by \(\binom{n + r - 1}{r - 1}\).
• Here, \(n = 3\) and \(r = 3\).
• \(\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10\).
• Distribution of ways accommodating all constraints = 11376
• Calculate the ways of distribution:

1. Conclusion:

The number of ways a student can attempt 15 questions satisfying all the given conditions is 11,376, making the correct option 11,376.