A candidate is required to answer 7 out of 15 questions which are divided into three groups A B C each containing 4 5 6 question respectively. He is required to select at least 2 questions from each group. He can make up his choice in

Options

A. 1200
B. 2700
C. 2000
D. none of these

Answer

To solve the problem of selecting 7 questions from 15, divided into three groups A, B, and C, we need to follow a systematic approach. Each group has a minimum requirement of selecting at least 2 questions. Here’s how we can break it down step by step:

### Step 1: Understand the Groups and Requirements
- Group A has 4 questions.
- Group B has 5 questions.
- Group C has 6 questions.
- We need to select a total of 7 questions with at least 2 from each group.

### Step 2: Define Cases Based on Selection
Since we need to select at least 2 questions from each group, we can define different cases based on how we distribute the remaining questions after selecting the minimum required from each group.

1. **Case 1**: Select 2 from A, 2 from B, and 3 from C.
2. **Case 2**: Select 2 from A, 3 from B, and 2 from C.
3. **Case 3**: Select 3 from A, 2 from B, and 2 from C.

### Step 3: Calculate Each Case

#### Case 1: 2 from A, 2 from B, 3 from C
- Number of ways to choose:
- From A: \( \binom{4}{2} \)
- From B: \( \binom{5}{2} \)
- From C: \( \binom{6}{3} \)

Calculating:
- \( \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \)
- \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)
- \( \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)

Total for Case 1:
\[ 6 \times 10 \times 20 = 1200 \]

#### Case 2: 2 from A, 3 from B, 2 from C
- Number of ways to choose:
- From A: \( \binom{4}{2} \)
- From B: \( \binom{5}{3} \)
- From C: \( \binom{6}{2} \)

Calculating:
- \( \binom{4}{2} = 6 \) (already calculated)
- \( \binom{5}{3} = \binom{5}{2} = 10 \) (since \( \binom{n}{r} = \binom{n}{n-r} \))
- \( \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \)

Total for Case 2:
\[ 6 \times 10 \times 15 = 900 \]

#### Case 3: 3 from A, 2 from B, 2 from C
- Number of ways to choose:
- From A: \( \binom{4}{3} \)
- From B: \( \binom{5}{2} \)
- From C: \( \binom{6}{2} \)

Calculating:
- \( \binom{4}{3} = 4 \)
- \( \binom{5}{2} = 10 \) (already calculated)
- \( \binom{6}{2} = 15 \) (already calculated)

Total for Case 3:
\[ 4 \times 10 \times 15 = 600 \]

### Step 4: Sum the Cases
Now, we sum the totals from all cases to find the total number of ways to select the questions:
\[ 1200 + 900 + 600 = 2700 \]

### Final Answer
The candidate can make his choice in **2700 ways**.

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