In how many ways can a student of XI choose 5 subjects out of 9 available subjects if two subjects, English (core) and Hindi (core) are compulsory.

Answer

To solve the problem of how many ways a student can choose 5 subjects out of 9 available subjects, given that English and Hindi are compulsory, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Subjects**:
- We have a total of 9 subjects.

2. **Identify the Compulsory Subjects**:
- The two compulsory subjects are English (core) and Hindi (core).

3. **Determine the Number of Subjects to Choose**:
- The student needs to choose 5 subjects in total.

4. **Calculate the Remaining Subjects to Choose**:
- Since 2 subjects (English and Hindi) are compulsory, the student needs to choose 3 more subjects.
- Remaining subjects to choose = Total subjects to choose - Compulsory subjects = 5 - 2 = 3.

5. **Calculate the Remaining Available Subjects**:
- After accounting for the compulsory subjects, the number of subjects left to choose from is 9 - 2 = 7 subjects.

6. **Use the Combination Formula**:
- We need to choose 3 subjects from the remaining 7 subjects. The number of ways to choose \( r \) subjects from \( n \) subjects is given by the combination formula:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
- Here, \( n = 7 \) and \( r = 3 \).

7. **Plug in the Values**:
\[ C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} \]

8. **Simplify the Factorials**:
- We can simplify \( 7! \) as \( 7 \times 6 \times 5 \times 4! \):
\[ C(7, 3) = \frac{7 \times 6 \times 5 \times 4!}{3! \cdot 4!} \]
- The \( 4! \) cancels out:
\[ C(7, 3) = \frac{7 \times 6 \times 5}{3!} \]

9. **Calculate \( 3! \)**:
- \( 3! = 3 \times 2 \times 1 = 6 \).

10. **Final Calculation**:
\[ C(7, 3) = \frac{7 \times 6 \times 5}{6} = 7 \times 5 = 35. \]

### Conclusion:
The number of ways a student can choose 5 subjects out of 9 available subjects, with English and Hindi being compulsory, is **35**.