Show the method :A committee of 5 is to be formed from 7 students and teachers. In how many ways can this be done if?
(a)The committee has 3 students and 2 teachers.
(b)The committee has 3 teachers and 2 students.
(c)There are no students on the committee.

To solve these problems, we'll use the concept of combinations, which is a way of selecting items from a larger group where the order does not matter. The formula for combinations is given by:

\[
\text{Number of combinations} = \binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

where:
- \( n \) is the total number of items to choose from.
- \( r \) is the number of items to be chosen.
- \( ! \) denotes factorial, which is the product of all positive integers up to that number.

Let's solve each part step by step.

### Part (a) The committee has 3 students and 2 teachers.

Here, we need to select 3 students from a group of 7 students and 2 teachers from a group of 7 teachers.

1. **Number of ways to select 3 students from 7 students**:
   \[
   \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
   \]

2. **Number of ways to select 2 teachers from 7 teachers**:
   \[
   \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21
   \]

3. **Total number of ways to form the committee**:
   \[
   \text{Total ways} = \binom{7}{3} \times \binom{7}{2} = 35 \times 21 = 735
   \]

### Part (b) The committee has 3 teachers and 2 students.

This is similar to part (a), but we reverse the roles of students and teachers.

1. **Number of ways to select 3 teachers from 7 teachers**:
   \[
   \binom{7}{3} = 35
   \]

2. **Number of ways to select 2 students from 7 students**:
   \[
   \binom{7}{2} = 21
   \]

3. **Total number of ways to form the committee**:
   \[
   \text{Total ways} = \binom{7}{3} \times \binom{7}{2} = 35 \times 21 = 735
   \]

### Part (c) There are no students on the committee.

In this case, the committee will consist entirely of teachers. We need to select 5 teachers from a group of 7 teachers.

1. **Number of ways to select 5 teachers from 7 teachers**:
   \[
   \binom{7}{5} = \binom{7}{2} = 21
   \]

Thus, the number of ways to form the committee when there are no students is 21.

### Summary of Results:
- (a) The committee with 3 students and 2 teachers can be formed in **735 ways**.
- (b) The committee with 3 teachers and 2 students can be formed in **735 ways**.
- (c) The committee with no students (only 5 teachers) can be formed in **21 ways**.

Would you like more details or have any questions?

### Related Questions:
1. How would the calculation change if there were 10 students and 8 teachers?
2. What if the committee must have exactly 4 students and 1 teacher?
3. How do we calculate the number of ways to form a committee with at least 2 students?
4. How many ways can the committee be formed if it must consist of all teachers or all students?
5. How does the inclusion of gender or other categories affect the calculation?

### Tip:
When dealing with combinations, remember that \( \binom{n}{r} = \binom{n}{n-r} \). This symmetry can simplify calculations!

Explore how to form committees from a group of students and teachers using combinations. Solve specific scenarios where committees consist of 3 students and 2 teachers, 3 teachers and 2 students, and exclusively teachers. Learn the combinatorial formula and its application in practical scenarios.

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