In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : A, B and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is __________.

Solution

Maths sir · 3 steps

Fix how many come from each section first, then count each split and add the cases.

1. Set up the split. Let a, b, c be the numbers taken from A, B, C. We need a + b + c = 15 with each at least 4, and a ≤ 8, b ≤ 6, c ≤ 6.
2. List the cases. The minimum 4 from each section uses 12 questions, so 3 extra questions are spread over the sections. B and C can take at most 2 extra each, because they have only 6 questions. That leaves 7 valid splits, and each split gives C(8, a)C(6, b)C(6, c) ways.

   (a, b, c)	Product	Ways
   (7, 4, 4)	8 × 15 × 15	1800
   (6, 5, 4)	28 × 6 × 15	2520
   (6, 4, 5)	28 × 15 × 6	2520
   (5, 5, 5)	56 × 6 × 6	2016
   (5, 6, 4) and (5, 4, 6)	56 × 1 × 15 each	840 + 840 = 1680
   (4, 6, 5) and (4, 5, 6)	70 × 1 × 6 each	420 + 420 = 840

3. So. Total = 1800 + 2520 + 2520 + 2016 + 1680 + 840 = 11376.

The common slip: 882000. Choosing 4 from each section first and then any 3 of the remaining 8 gives 70 × 15 × 15 × 56 = 882000. This counts the same set of 15 questions many times.

The idea to keep. With an 'at least' condition on each group, list every valid split of the total, multiply the choices within each split, then add. Always check the upper limits of each group too.

Answer from the NTA final answer key. Solution by PrepWiser.