In how many ways can 5 subjects be chosen from 9 subjects if three subjects are compulsory?

Answer

To solve the problem of selecting 5 subjects from 9 subjects when 3 subjects are compulsory, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compulsory Subjects**:
Since 3 subjects are compulsory, we will always include these 3 subjects in our selection.

2. **Determine Remaining Subjects**:
After selecting the 3 compulsory subjects from the total of 9 subjects, we have:
\[ 9 - 3 = 6 \text{ subjects remaining.} \]

3. **Calculate the Number of Subjects to Choose**:
We need to select a total of 5 subjects. Since we have already chosen 3 compulsory subjects, we need to choose:
\[ 5 - 3 = 2 \text{ subjects from the remaining 6 subjects.} \]

4. **Use the Combination Formula**:
The number of ways to choose 2 subjects from 6 subjects can be calculated using the combination formula:
\[ nCr = \frac{n!}{r!(n - r)!} \]
Here, \( n = 6 \) and \( r = 2 \). So we need to calculate:
\[ 6C2 = \frac{6!}{2!(6 - 2)!} = \frac{6!}{2! \cdot 4!} \]

5. **Simplify the Factorials**:
We can simplify \( 6! \) as follows:
\[ 6! = 6 \times 5 \times 4! \]
Thus, substituting this back into our combination formula gives:
\[ 6C2 = \frac{6 \times 5 \times 4!}{2! \times 4!} \]
The \( 4! \) terms cancel out:
\[ 6C2 = \frac{6 \times 5}{2!} \]

6. **Calculate \( 2! \)**:
We know that \( 2! = 2 \), so:
\[ 6C2 = \frac{6 \times 5}{2} = \frac{30}{2} = 15 \]

7. **Final Answer**:
Therefore, the total number of ways to choose 5 subjects from 9 subjects, with 3 subjects being compulsory, is:
\[ \boxed{15} \]