A committee of five members is to be formed out of Males , 6 Females and 3 children ,In how many different ways can it be done if -?
The committee should consist of 2 Males, 2Females and I Child ?

Options

A. 450
B. 225
C. 55
D. 90

Explanation

To solve the problem of forming a committee of 5 members consisting of 2 males, 2 females, and 1 child from the available members, we will use the concept of combinations.

### Step-by-Step Solution:

1. **Identify the Available Members**:
- Males: 5
- Females: 6
- Children: 3

2. **Determine the Required Committee Composition**:
- Males: 2
- Females: 2
- Children: 1

3. **Calculate the Number of Ways to Choose Males**:
We need to choose 2 males from 5 available males. The number of ways to choose 2 males from 5 can be calculated using the combination formula:
\[ \text{Number of ways to choose 2 males} = \binom{5}{2} \]
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

4. **Calculate the Number of Ways to Choose Females**:
We need to choose 2 females from 6 available females. The number of ways to choose 2 females from 6 is:
\[ \text{Number of ways to choose 2 females} = \binom{6}{2} \]
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

5. **Calculate the Number of Ways to Choose Children**:
We need to choose 1 child from 3 available children. The number of ways to choose 1 child from 3 is:
\[ \text{Number of ways to choose 1 child} = \binom{3}{1} \]
\[ \binom{3}{1} = \frac{3!}{1!(3-1)!} = 3 \]

6. **Combine the Results**:
To find the total number of ways to form the committee, we multiply the number of ways to choose males, females, and children:
\[ \text{Total ways} = \binom{5}{2} \times \binom{6}{2} \times \binom{3}{1} \]
\[ \text{Total ways} = 10 \times 15 \times 3 = 450 \]

### Final Answer:
The total number of different ways to form the committee is **450**.