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 each group. The number of ways in which he can choose the 7 questions is

Options

A. (a)780
B. (b)640
C. (c)820
D. (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 (each containing 6 questions) with the restriction that no more than 5 questions can be chosen from each group, we can break it down into cases based on how many questions are chosen from each group.

### Step-by-Step Solution:

1. **Identify the Groups**:
- Let Group 1 have questions \( Q_1, Q_2, Q_3, Q_4, Q_5, Q_6 \).
- Let Group 2 have questions \( Q_7, Q_8, Q_9, Q_{10}, Q_{11}, Q_{12} \).

2. **Define Cases**:
- Since the candidate can choose a maximum of 5 questions from each group, we can define the following cases based on the number of questions chosen from Group 1 (denote it as \( x \)) and from Group 2 (denote it as \( y \)):
- Case 1: \( x = 5 \) and \( y = 2 \)
- Case 2: \( x = 4 \) and \( y = 3 \)
- Case 3: \( x = 3 \) and \( y = 4 \)
- Case 4: \( x = 2 \) and \( y = 5 \)

3. **Calculate Each Case**:
- **Case 1**: \( x = 5 \), \( y = 2 \)
\[ \text{Ways} = \binom{6}{5} \times \binom{6}{2} = 6 \times 15 = 90 \]

- **Case 2**: \( x = 4 \), \( y = 3 \)
\[ \text{Ways} = \binom{6}{4} \times \binom{6}{3} = 15 \times 20 = 300 \]

- **Case 3**: \( x = 3 \), \( y = 4 \)
\[ \text{Ways} = \binom{6}{3} \times \binom{6}{4} = 20 \times 15 = 300 \]

- **Case 4**: \( x = 2 \), \( y = 5 \)
\[ \text{Ways} = \binom{6}{2} \times \binom{6}{5} = 15 \times 6 = 90 \]

4. **Sum the Cases**:
- Now, we add the number of ways from all cases:
\[ \text{Total Ways} = 90 + 300 + 300 + 90 = 780 \]

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

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