In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time?

Answer

To solve the problem of forming a committee of five persons from eight members, where a particular member is included every time, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Members and the Fixed Member**:
- We have a total of 8 members.
- One particular member is included in every committee. Let's call this member A.

2. **Determine Remaining Members to Choose From**:
- Since member A is already included in the committee, we only need to choose the remaining members from the other 7 members (B, C, D, E, F, G, H).

3. **Calculate the Number of Members to Choose**:
- We need to choose 4 additional members from the remaining 7 members (B, C, D, E, F, G, H).

4. **Use the Combination Formula**:
- The number of ways to choose r members from n members is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
- In our case, we need to calculate \( \binom{7}{4} \).

5. **Substitute Values into the Combination Formula**:
- Here, \( n = 7 \) and \( r = 4 \):
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} \]

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

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

8. **Final Calculation**:
- Now substitute \( 3! \) back into the equation:
\[ \binom{7}{4} = \frac{7 \times 6 \times 5}{6} \]
- The 6 cancels out:
\[ = 7 \times 5 = 35 \]

9. **Conclusion**:
- Therefore, the number of ways to form the committee is **35**.