In how many ways can a committee of 4 be selected out of 12 persons so that a particular person may
always be taken.

Answer

To solve the problem of selecting a committee of 4 persons from a total of 12, with the condition that a particular person must always be included, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Persons and the Particular Person**:
- We have a total of 12 persons.
- Let's denote the particular person who must always be included as Person A.

2. **Select the Remaining Members**:
- Since Person A is always included in the committee, we need to select 3 more members from the remaining 11 persons (12 total - 1 particular person = 11 remaining).

3. **Use the Combination Formula**:
- The number of ways to choose r persons from n persons is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
- In our case, we need to calculate:
\[ \binom{11}{3} \]
- Here, \( n = 11 \) (the remaining persons) and \( r = 3 \) (the number of additional members to select).

4. **Calculate \( \binom{11}{3} \)**:
- Substitute the values into the formula:
\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3! \cdot 8!} \]
- We can simplify this by expanding \( 11! \):
\[ \binom{11}{3} = \frac{11 \times 10 \times 9 \times 8!}{3! \times 8!} \]
- The \( 8! \) cancels out:
\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3!} \]

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

6. **Final Calculation**:
- Now substitute \( 3! \) back into the equation:
\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{6} \]
- Calculate the numerator:
\[ 11 \times 10 = 110 \]
\[ 110 \times 9 = 990 \]
- Now divide by 6:
\[ \frac{990}{6} = 165 \]

7. **Conclusion**:
- Therefore, the total number of ways to select a committee of 4 persons from 12, ensuring that a particular person is always included, is **165**.