In how many ways can a committee of 6 members e formed out of 4 teachers and 7 students when
(i) one teacher is in the committee?
(ii) at least one teacher is in the committee?

Answer

To solve the problem of forming a committee of 6 members from 4 teachers and 7 students under the given conditions, we will break it down into two parts.

### Part (i): One Teacher in the Committee

1. **Select 1 Teacher from 4 Teachers:**
We can choose 1 teacher from the 4 available teachers. The number of ways to choose 1 teacher is given by the combination formula:
\[ \binom{4}{1} = 4 \]

2. **Select 5 Students from 7 Students:**
After selecting 1 teacher, we need to select 5 students from the 7 available students. The number of ways to choose 5 students is:
\[ \binom{7}{5} = \binom{7}{2} = \frac{7!}{5! \cdot 2!} = \frac{7 \times 6}{2 \times 1} = 21 \]

3. **Calculate Total Ways:**
The total number of ways to form the committee with exactly 1 teacher and 5 students is the product of the two combinations calculated:
\[ \text{Total Ways} = \binom{4}{1} \times \binom{7}{5} = 4 \times 21 = 84 \]

### Part (ii): At Least One Teacher in the Committee

To find the number of ways to form a committee with at least one teacher, we can consider all possible cases where the number of teachers varies from 1 to 4.

1. **Case 1: 1 Teacher and 5 Students**
- Already calculated: \( \binom{4}{1} \times \binom{7}{5} = 4 \times 21 = 84 \)

2. **Case 2: 2 Teachers and 4 Students**
- Choose 2 teachers from 4:
\[ \binom{4}{2} = 6 \]
- Choose 4 students from 7:
\[ \binom{7}{4} = 35 \]
- Total for this case:
\[ 6 \times 35 = 210 \]

3. **Case 3: 3 Teachers and 3 Students**
- Choose 3 teachers from 4:
\[ \binom{4}{3} = 4 \]
- Choose 3 students from 7:
\[ \binom{7}{3} = 35 \]
- Total for this case:
\[ 4 \times 35 = 140 \]

4. **Case 4: 4 Teachers and 2 Students**
- Choose 4 teachers from 4:
\[ \binom{4}{4} = 1 \]
- Choose 2 students from 7:
\[ \binom{7}{2} = 21 \]
- Total for this case:
\[ 1 \times 21 = 21 \]

5. **Calculate Total Ways for At Least One Teacher:**
Now, we sum the totals from all cases:
\[ \text{Total} = 84 + 210 + 140 + 21 = 455 \]

### Final Answers:
- (i) The number of ways to form a committee with one teacher is **84**.
- (ii) The number of ways to form a committee with at least one teacher is **455**.