In how many ways can 5 members form a committee out of 10 be selected so that
(i) two particular members must be included.
(ii) two particular members must not be included.

Answer

To solve the problem of forming a committee of 5 members from a group of 10 members with specific conditions, we will break it down into two parts as given in the question.

### Part (i): Two particular members must be included.

1. **Identify the fixed members**: Since two particular members must be included in the committee, we will consider these two members as already selected.

2. **Determine the remaining members**: With 2 members already included, we have to select the remaining members from the remaining pool. Since there are 10 members total and 2 are already chosen, we have 10 - 2 = 8 members left.

3. **Calculate the number of ways to choose the remaining members**: We need to select 3 more members from the remaining 8 members. The number of ways to choose 3 members from 8 can be calculated using the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Here, \( n = 8 \) and \( r = 3 \):
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

4. **Conclusion for part (i)**: Therefore, the total number of ways to form a committee of 5 members including the 2 particular members is **56**.

### Part (ii): Two particular members must not be included.

1. **Identify the excluded members**: Since two particular members must not be included in the committee, we will exclude these two members from our selection.

2. **Determine the remaining members**: With 2 members excluded, we have 10 - 2 = 8 members left to choose from.

3. **Calculate the number of ways to choose the committee**: We need to select all 5 members from these remaining 8 members. The number of ways to choose 5 members from 8 can also be calculated using the combination formula:
\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

4. **Conclusion for part (ii)**: Therefore, the total number of ways to form a committee of 5 members excluding the 2 particular members is also **56**.

### Final Answers:
- (i) 56 ways when 2 particular members must be included.
- (ii) 56 ways when 2 particular members must not be included.