In an examination, Yamini has to select 4 questions from each part. There are 6, 7 and 8 questions in Part I, Part II and Part III respectively. What is the number of possible combinations in which she can choose the questions ?

Answer

To solve the problem, we need to calculate the number of ways Yamini can select 4 questions from each of the three parts of the examination. We will use the combination formula \( nCr = \frac{n!}{r!(n-r)!} \).

### Step-by-Step Solution:

1. **Identify the number of questions in each part:**
- Part I has 6 questions.
- Part II has 7 questions.
- Part III has 8 questions.

2. **Determine the combinations for each part:**
- For Part I, the number of ways to choose 4 questions from 6 is given by \( 6C4 \).
- For Part II, the number of ways to choose 4 questions from 7 is given by \( 7C4 \).
- For Part III, the number of ways to choose 4 questions from 8 is given by \( 8C4 \).

3. **Calculate each combination:**
- **For Part I:**
\[ 6C4 = \frac{6!}{4!(6-4)!} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15 \]

- **For Part II:**
\[ 7C4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

- **For Part III:**
\[ 8C4 = \frac{8!}{4!(8-4)!} = \frac{8!}{4! \cdot 4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

4. **Multiply the combinations from all parts:**
\[ \text{Total combinations} = 6C4 \times 7C4 \times 8C4 = 15 \times 35 \times 70 \]

5. **Calculate the total combinations:**
- First, calculate \( 15 \times 35 \):
\[ 15 \times 35 = 525 \]
- Next, multiply the result by 70:
\[ 525 \times 70 = 36750 \]

6. **Final Answer:**
The total number of possible combinations in which Yamini can choose the questions is **36750**.