A question paper has three sections A,B, C having 8, 6, 4 questions respectively. If the student has to answer 15 questions attempting atleast four from each section. Find the number of ways the paper can be answered by student.

Options

A. 342
B. 344
C. 374
D. 340

Answer

To solve the problem of how many ways a student can answer a question paper with three sections (A, B, C) while adhering to the given constraints, we can break down the solution step by step.

### Step 1: Understand the Problem
We have three sections:
- Section A: 8 questions
- Section B: 6 questions
- Section C: 4 questions

The student must answer a total of 15 questions, with at least 4 questions from each section.

### Step 2: Set Up the Variables
Let:
- \( x \): number of questions answered from Section A
- \( y \): number of questions answered from Section B
- \( z \): number of questions answered from Section C

We know:
1. \( x + y + z = 15 \)
2. \( x \geq 4 \)
3. \( y \geq 4 \)
4. \( z \geq 4 \)

### Step 3: Substitute the Minimum Values
Since the student must answer at least 4 questions from each section, we can substitute:
- Let \( x' = x - 4 \) (so \( x' \geq 0 \))
- Let \( y' = y - 4 \) (so \( y' \geq 0 \))
- Let \( z' = z - 4 \) (so \( z' \geq 0 \))

Now, we can rewrite the equation:
\[ (x' + 4) + (y' + 4) + (z' + 4) = 15 \]
This simplifies to:
\[ x' + y' + z' = 3 \]

### Step 4: Find Non-Negative Integer Solutions
We need to find the number of non-negative integer solutions to the equation \( x' + y' + z' = 3 \). This can be solved using the "stars and bars" combinatorial method.

The formula for the number of solutions is given by:
\[ \binom{n+k-1}{k-1} \]
where \( n \) is the total number of items to distribute (3 in our case) and \( k \) is the number of groups (3 sections).

Thus, we calculate:
\[ \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} = 10 \]

### Step 5: Calculate the Combinations for Each Case
Now we will consider the combinations for each case of \( (x, y, z) \) based on the number of questions chosen from each section.

1. **Case 1**: \( (x, y, z) = (7, 4, 4) \)
- Ways to choose: \( \binom{8}{7} \cdot \binom{6}{4} \cdot \binom{4}{4} = 8 \cdot 15 \cdot 1 = 120 \)

2. **Case 2**: \( (x, y, z) = (6, 5, 4) \)
- Ways to choose: \( \binom{8}{6} \cdot \binom{6}{5} \cdot \binom{4}{4} = 28 \cdot 6 \cdot 1 = 168 \)

3. **Case 3**: \( (x, y, z) = (6, 4, 5) \)
- Ways to choose: \( \binom{8}{6} \cdot \binom{6}{4} \cdot \binom{4}{4} = 28 \cdot 15 \cdot 1 = 420 \)

4. **Case 4**: \( (x, y, z) = (5, 6, 4) \)
- Ways to choose: \( \binom{8}{5} \cdot \binom{6}{6} \cdot \binom{4}{4} = 56 \cdot 1 \cdot 1 = 56 \)

### Step 6: Total the Cases
Now, we sum all the cases:
\[ 120 + 168 + 56 + 420 = 764 \]

### Final Answer
Thus, the total number of ways the student can answer the paper is **764**.

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