A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. The number of ways in which he can choose the 7 questions is

Options

A. 650
B. 780
C. 720
D. none of these

Answer

To solve the problem of how many ways a candidate can choose 7 questions from 12 questions divided into two groups (A and B), with each group containing 6 questions and the restriction of not attempting more than 5 questions from either group, we can break down the solution as follows:

### Step-by-Step Solution

1. **Identify the Groups**:
- Group A: 6 questions
- Group B: 6 questions

2. **Understand the Constraints**:
- The candidate must select a total of 7 questions.
- The candidate cannot select more than 5 questions from either group.

3. **Determine Possible Cases**:
- Since the candidate can select up to 5 questions from either group, we can have the following cases:
- Case 1: 5 questions from Group A and 2 questions from Group B
- Case 2: 4 questions from Group A and 3 questions from Group B
- Case 3: 3 questions from Group A and 4 questions from Group B
- Case 4: 2 questions from Group A and 5 questions from Group B

4. **Calculate the Number of Ways for Each Case**:
- **Case 1**: Selecting 5 from A and 2 from B
\[ \text{Ways} = \binom{6}{5} \times \binom{6}{2} = 6 \times 15 = 90 \]
- **Case 2**: Selecting 4 from A and 3 from B
\[ \text{Ways} = \binom{6}{4} \times \binom{6}{3} = 15 \times 20 = 300 \]
- **Case 3**: Selecting 3 from A and 4 from B
\[ \text{Ways} = \binom{6}{3} \times \binom{6}{4} = 20 \times 15 = 300 \]
- **Case 4**: Selecting 2 from A and 5 from B
\[ \text{Ways} = \binom{6}{2} \times \binom{6}{5} = 15 \times 6 = 90 \]

5. **Sum the Ways from All Cases**:
\[ \text{Total Ways} = 90 + 300 + 300 + 90 = 780 \]

### Final Answer
The total number of ways the candidate can choose 7 questions is **780**.

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