In how many ways can 5 members forming 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 selecting 5 members from a committee of 10 with specific conditions, we will break it down into two parts as stated in the question.

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

1. **Identify the total members and the requirement**: We have a total of 10 members, and we need to select 5 members. Out of these, 2 particular members must be included in the selection.

2. **Determine how many members are left to choose**: Since 2 members are already included, we need to select 3 more members from the remaining members.

\[ \text{Remaining members} = 10 - 2 = 8 \]

3. **Calculate the number of ways to select the remaining members**: We need to choose 3 members from the 8 remaining members. This can be calculated using the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Here, \( n = 8 \) and \( r = 3 \):

\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} \]

4. **Simplify the combination**:

\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

Thus, the number of ways to select the committee when 2 particular members must be included is **56**.

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

1. **Identify the total members and the requirement**: Again, we have 10 members, but this time, 2 particular members must not be included in the selection.

2. **Determine how many members are left to choose**: Since 2 members are excluded, we are left with:

\[ \text{Remaining members} = 10 - 2 = 8 \]

3. **Calculate the number of ways to select the committee**: We need to select all 5 members from the 8 remaining members. This can be calculated using the combination formula:

\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5! \cdot 3!} \]

4. **Simplify the combination**:

\[ \binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

Thus, the number of ways to select the committee when 2 particular members must not be included is also **56**.

### Summary of Results:
- (i) Number of ways with 2 particular members included: **56**
- (ii) Number of ways with 2 particular members excluded: **56**