How many committees of 5 members each can be formed with 8 officials and non - official members in the following cases:
a particular non-official members is always included?

Answer

To solve the problem of how many committees of 5 members can be formed with 8 officials and non-official members, given that a particular non-official member is always included, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Members**:
We have 8 officials and non-official members. This means we have a total of 8 members to choose from.

2. **Include the Particular Non-Official Member**:
Since one particular non-official member must always be included in the committee, we can consider this member as already selected. This leaves us needing to select 4 more members.

3. **Calculate the Remaining Members**:
After including the particular non-official member, we have 7 members left (8 total - 1 selected = 7 remaining).

4. **Select 4 Members from the Remaining 7**:
We now need to choose 4 members from the remaining 7 members. The number of ways to do this can be calculated using the combination formula \( nCr = \frac{n!}{r!(n-r)!} \).

5. **Apply the Combination Formula**:
Here, \( n = 7 \) and \( r = 4 \):
\[ \text{Number of ways} = \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} \]

6. **Calculate Factorials**:
\[ 7! = 7 \times 6 \times 5 \times 4! \quad \text{(we can cancel } 4! \text{ later)} \]
\[ 3! = 3 \times 2 \times 1 = 6 \]

7. **Simplify the Expression**:
\[ \binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 \]

8. **Final Answer**:
Therefore, the total number of committees of 5 members that can be formed, with the particular non-official member included, is **35**.