An examination paper consists of 12 questions divided into two parts, part A and part 8. Part A contains 7 questions and part B contains 5 questions. A candidate is required to attempt 8 questions, selecting atleast 3 from each part. In how many ways can he select the questions?

Options

A. 240
B. 60
C. 420
D. 480

Answer

To solve the problem of how many ways a candidate can select 8 questions from an examination paper consisting of 12 questions divided into two parts (Part A with 7 questions and Part B with 5 questions), while ensuring that at least 3 questions are selected from each part, we can break down the solution into several steps.

### Step-by-Step Solution:

1. **Understanding the Selection Criteria**:
- The candidate must select a total of 8 questions.
- At least 3 questions must be selected from Part A (7 questions) and at least 3 questions from Part B (5 questions).

2. **Identifying Possible Cases**:
- Since the candidate must select at least 3 questions from each part, we can outline the possible distributions of questions selected from each part:
- Case 1: 3 from Part A and 5 from Part B
- Case 2: 4 from Part A and 4 from Part B
- Case 3: 5 from Part A and 3 from Part B

3. **Calculating the Number of Ways for Each Case**:
- **Case 1**: Selecting 3 from Part A and 5 from Part B
- Number of ways to choose 3 questions from Part A: \( \binom{7}{3} \)
- Number of ways to choose 5 questions from Part B: \( \binom{5}{5} \)
- Total ways for Case 1:
\[ \binom{7}{3} \times \binom{5}{5} = 35 \times 1 = 35 \]

- **Case 2**: Selecting 4 from Part A and 4 from Part B
- Number of ways to choose 4 questions from Part A: \( \binom{7}{4} \)
- Number of ways to choose 4 questions from Part B: \( \binom{5}{4} \)
- Total ways for Case 2:
\[ \binom{7}{4} \times \binom{5}{4} = 35 \times 5 = 175 \]

- **Case 3**: Selecting 5 from Part A and 3 from Part B
- Number of ways to choose 5 questions from Part A: \( \binom{7}{5} \)
- Number of ways to choose 3 questions from Part B: \( \binom{5}{3} \)
- Total ways for Case 3:
\[ \binom{7}{5} \times \binom{5}{3} = 21 \times 10 = 210 \]

4. **Summing the Total Ways**:
- Now, we will add the total ways from all the cases:
\[ 35 + 175 + 210 = 420 \]

5. **Conclusion**:
- The total number of ways a candidate can select the questions is **420**.