Question

A student has to answer 10 questions, choosing at least 4 from each of the parts A and B. If there are 6 questions in part A and 7 in part B, then the number of ways can the student choose 10 questions is

• A. 256
• B. 352
• C. 266
• D. 426

Answer 1

The Correct Option is C

The student needs to choose 10 questions, at least 4 from part A and 4 from part B. Let’s break down the selection process:

• There are 6 questions in part A and 7 questions in part B.
• The student must choose at least 4 questions from part A and at least 4 questions from part B. Therefore, the student must choose either 4, 5, or 6 questions from part A, and the remaining questions will be from part B.

We will now calculate the number of ways the student can choose questions:
1. Choose 4 questions from part A and 6 questions from part B: \[ \text{Ways} = \binom{6}{4} \times \binom{7}{6} = 15 \times 7 = 105 \]
2. Choose 5 questions from part A and 5 questions from part B: \[ \text{Ways} = \binom{6}{5} \times \binom{7}{5} = 6 \times 21 = 126 \]
3. Choose 6 questions from part A and 4 questions from part B: \[ \text{Ways} = \binom{6}{6} \times \binom{7}{4} = 1 \times 35 = 35 \] Now, add up all the possibilities: \[ \text{Total ways} = 105 + 126 + 35 = 266 \]

The correct answer is (C) : 266.

Answer 2

The student has to answer 10 questions in total, choosing at least 4 from each part. Part A has 6 questions and part B has 7 questions.

We need to find the number of ways the student can choose the questions.

Let's consider the possible distributions of questions from Part A and Part B:

• Case 1: 4 questions from Part A and 6 questions from Part B. The number of ways to choose 4 questions from Part A is \(\binom{6}{4}\). The number of ways to choose 6 questions from Part B is \(\binom{7}{6}\). So, the number of ways for this case is \(\binom{6}{4} \times \binom{7}{6} = \frac{6!}{4!2!} \times \frac{7!}{6!1!} = 15 \times 7 = 105\).
• Case 2: 5 questions from Part A and 5 questions from Part B. The number of ways to choose 5 questions from Part A is \(\binom{6}{5}\). The number of ways to choose 5 questions from Part B is \(\binom{7}{5}\). So, the number of ways for this case is \(\binom{6}{5} \times \binom{7}{5} = \frac{6!}{5!1!} \times \frac{7!}{5!2!} = 6 \times 21 = 126\).
• Case 3: 6 questions from Part A and 4 questions from Part B. The number of ways to choose 6 questions from Part A is \(\binom{6}{6}\). The number of ways to choose 4 questions from Part B is \(\binom{7}{4}\). So, the number of ways for this case is \(\binom{6}{6} \times \binom{7}{4} = \frac{6!}{6!0!} \times \frac{7!}{4!3!} = 1 \times 35 = 35\).

The total number of ways is the sum of the ways for each case: \(105 + 126 + 35 = 266\).

Therefore, the number of ways the student can choose 10 questions is 266.