A committee of 11 is to be formed from 8 teachers and 12 students of whom 4 are girls. In how many ways this can be done so that the committee contains at least four of either groups (teachers and students) and at least 2 girls and atleast 2 boys are in the committee.

Answer

To solve the problem of forming a committee of 11 members from 8 teachers and 12 students (of which 4 are girls), while ensuring that the committee contains at least 4 members from either group and at least 2 girls and 2 boys, we can break down the solution into several cases.

### Step-by-Step Solution:

1. **Identify the Groups**:
- Teachers: 8
- Students: 12 (4 girls and 8 boys)

2. **Define the Conditions**:
- At least 4 members from either group (teachers or students).
- At least 2 girls and at least 2 boys in the committee.

3. **Consider Different Cases**:
We can form the committee in the following cases based on the number of teachers and students:

**Case 1**: 4 Teachers and 7 Students
**Case 2**: 5 Teachers and 6 Students
**Case 3**: 6 Teachers and 5 Students
**Case 4**: 7 Teachers and 4 Students

4. **Calculate Each Case**:

**Case 1**: 4 Teachers and 7 Students
- Possible distributions of girls and boys among 7 students:
- 4 Girls and 3 Boys:
\[ \binom{4}{4} \cdot \binom{8}{3} \]
- 3 Girls and 4 Boys:
\[ \binom{4}{3} \cdot \binom{8}{4} \]
- 2 Girls and 5 Boys:
\[ \binom{4}{2} \cdot \binom{8}{5} \]
- Total for Case 1:
\[ \binom{8}{4} \cdot \left( \binom{4}{4} \cdot \binom{8}{3} + \binom{4}{3} \cdot \binom{8}{4} + \binom{4}{2} \cdot \binom{8}{5} \right) \]

**Case 2**: 5 Teachers and 6 Students
- Possible distributions of girls and boys among 6 students:
- 4 Girls and 2 Boys:
\[ \binom{4}{4} \cdot \binom{8}{2} \]
- 3 Girls and 3 Boys:
\[ \binom{4}{3} \cdot \binom{8}{3} \]
- 2 Girls and 4 Boys:
\[ \binom{4}{2} \cdot \binom{8}{4} \]
- Total for Case 2:
\[ \binom{8}{5} \cdot \left( \binom{4}{4} \cdot \binom{8}{2} + \binom{4}{3} \cdot \binom{8}{3} + \binom{4}{2} \cdot \binom{8}{4} \right) \]

**Case 3**: 6 Teachers and 5 Students
- Possible distributions of girls and boys among 5 students:
- 3 Girls and 2 Boys:
\[ \binom{4}{3} \cdot \binom{8}{2} \]
- 2 Girls and 3 Boys:
\[ \binom{4}{2} \cdot \binom{8}{3} \]
- Total for Case 3:
\[ \binom{8}{6} \cdot \left( \binom{4}{3} \cdot \binom{8}{2} + \binom{4}{2} \cdot \binom{8}{3} \right) \]

**Case 4**: 7 Teachers and 4 Students
- Possible distribution of girls and boys among 4 students:
- 2 Girls and 2 Boys:
\[ \binom{4}{2} \cdot \binom{8}{2} \]
- Total for Case 4:
\[ \binom{8}{7} \cdot \left( \binom{4}{2} \cdot \binom{8}{2} \right) \]

5. **Calculate the Combinations**:
Now we can calculate each case:
- Case 1:
\[ \binom{8}{4} \cdot \left( 1 \cdot \binom{8}{3} + 4 \cdot \binom{8}{4} + 6 \cdot \binom{8}{5} \right) \]
- Case 2:
\[ \binom{8}{5} \cdot \left( 1 \cdot \binom{8}{2} + 4 \cdot \binom{8}{3} + 6 \cdot \binom{8}{4} \right) \]
- Case 3:
\[ \binom{8}{6} \cdot \left( 4 \cdot \binom{8}{2} + 6 \cdot \binom{8}{3} \right) \]
- Case 4:
\[ \binom{8}{7} \cdot \left( 6 \cdot \binom{8}{2} \right) \]

6. **Sum All Cases**:
Finally, sum the results from all four cases to get the total number of ways to form the committee.